Divisibility Rules Problems with Solutions

Divisibility Rules of Binomial Expansion

Edited By Komal Miglani | Updated on Jul 02, 2025 08:01 PM IST

An algebraic expression consisting of only two terms is called a Binomial Expression. Expressions with higher powers are difficult to solve. In these cases, Binomial theorem plays a major role. Binomial theorem is used for the expansion of a binomial expression with a higher degree. Binomial theorem is proved using the concept of mathematical induction. Apart from mathematics, Binomial theorem is also used in various fields for statistical and financial data analysis.

Divisibility Rules of Binomial Expansion

This article is about the divisibility rules of binomial expansion which falls under the topic Binomial theorem and its applications. This is one of the important topics for competitive exams.

Binomial Theorem

If $n$ is any positive integer, then

$ (a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{n - 1} b + \binom{n}{2} a^{n - 2} b^2 + \dots + \binom{n}{n - 1} a b^{n - 1} + \binom{n}{n} b^n $

Special cases of Binomial Theorem:

In the expansion of $ (a + b)^n $,

(i) Taking $ a = x $ and $ b = -y $, we obtain:

$(x - y)^n = [x + (-y)]^n $

$= \binom{n}{0} x^n + \binom{n}{1} x^{n - 1} (-y) + \binom{n}{2} x^{n - 2} (-y)^2 + \binom{n}{3} x^{n - 3} (-y)^3 + \dots + \binom{n}{n} (-y)^n $

$= \binom{n}{0} x^n - \binom{n}{1} x^{n - 1} y + \binom{n}{2} x^{n - 2} y^2 - \binom{n}{3} x^{n - 3} y^3 + \dots + (-1)^n \binom{n}{n} y^n $

Thus,

$(x - y)^n = \binom{n}{0} x^n - \binom{n}{1} x^{n - 1} y + \binom{n}{2} x^{n - 2} y^2 + \dots + (-1)^n \binom{n}{n} y^n$

Using this, we have:

$ (x - 2y)^5 = \binom{5}{0} x^5 - \binom{5}{1} x^4 (2y) + \binom{5}{2} x^3 (2y)^2 - \binom{5}{3} x^2 (2y)^3 + \binom{5}{4} x (2y)^4 - \binom{5}{5} (2y)^5 $

$= x^5 - 10 x^4 y + 40 x^3 y^2 - 80 x^2 y^3 + 80 x y^4 - 32 y^5 $

(ii) Taking $ a = 1 $, $ b = x $, we obtain:

$ (1 + x)^n = \binom{n}{0} (1)^n + \binom{n}{1} (1)^{n - 1} x + \binom{n}{2} (1)^{n - 2} x^2 + \dots + \binom{n}{n} x^n $

$ = \binom{n}{0} + \binom{n}{1} x + \binom{n}{2} x^2 + \binom{n}{3} x^3 + \dots + \binom{n}{n} x^n $

Thus,

$ (1 + x)^n = \binom{n}{0} + \binom{n}{1} x + \binom{n}{2} x^2 + \binom{n}{3} x^3 + \dots + \binom{n}{n} x^n $

In particular, for $ x = 1 $, we have:

$ 2^n = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} $

(iii) Taking $ a = 1 $, $ b = -x $, we obtain:

$ (1 - x)^n = \binom{n}{0} - \binom{n}{1} x + \binom{n}{2} x^2 - \dots + (-1)^n \binom{n}{n} x^n $

In particular, for $ x = 1 $, we get:

$ 0 = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \dots + (-1)^n \binom{n}{n} $

Divisibility Rules:

To find whether the binomial expansion is divisible by an integer,

1. Expression, $(1+x)^{\mathrm{n}}-1$ is divisible by x because

$ (1+\mathrm{x})^{\mathrm{n}}-1 =\left({ }^{\mathrm{n}} \mathrm{C}_0+{ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}+{ }^{\mathrm{n}} C_2 x^2+\ldots . .+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}\right)-1 $

$ =\mathrm{x}\left[{ }^{\mathrm{n}} \mathrm{C}_1+{ }^{\mathrm{n}} C_2 \mathrm{x}+\ldots \ldots .+{ }^{\mathrm{n}} C_{\mathrm{n}} \mathrm{x}^{\mathrm{n}-1}\right] $

2. Expression, $(1+\mathrm{x})^{\mathrm{n}}-\mathrm{nx}-1$ is divisible by $\mathrm{x}^2$ because

$ (1+\mathrm{x})^{\mathrm{n}}-\mathrm{nx}-1=\left({ }^{\mathrm{n}} \mathrm{C}_0+{ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+\ldots \ldots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}\right)-\mathrm{nx}-1 $

$ =x^2\left[{ }^n C_2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}+\ldots \ldots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}-2}\right] $

For example:

Prove that $3^{2 n+2}-8 n-9$ is divisible by 8 if $n \in N$

$ 3^{2 n+2} -8 n-9=(1+8)^{n+1}-8 n-9 $

$ =\left[1+(n+1) 8+{ }^{n+1} C_2 8^2+\ldots\right]-8 n-9$

$ ={ }^{n+1} C_2 8^2+{ }^{n+1} C_3 8^3+{ }^{n+1} C_4 8^4+\ldots $

$ =8\left[{ }^{n+1} C_2 8+{ }^{n+1} C_3 8^2+{ }^{n+1} C_4 8^3+\ldots\right] $

which is clearly divisible by 8 .

1. The expression $a^{\mathrm{n}}-\mathrm{b}^{\mathrm{n}}$ is divisible by $\mathrm{a}+\mathrm{b}$, if n is even.

2. The expression $a^n-b^n$ is divisible by $a-b$, if $n$ is even or odd.

3. The expression $a^{\mathrm{n}}+b^{\mathrm{n}^i}$ divisible by, $\mathrm{a}+$ bif n is odd.

In all the above cases $n$ is a natural number.

For Example,

The expression $15^4-7^4$ is divisible by $(15+7)=22$ and $(15-7)=8$

Solved Examples based on divisibility rules of binomial expansion:

Example 1: Which of the following is true for the number $2^{4 n}-2^n(7 n+1)$, where n is any positive Integer?

1) The highest divisor of the number among the given options is $7$.

2) The highest divisor of the number among the given options is $72$.

3) The highest divisor of the number among the given options is $142$.

4) The highest divisor of the number among the given options is $14$.

Solution:

$ 2^{4 n}-2^n(7 n+1)=(16)^n-2^n(7 n+1) $

$ \quad=(2+14)^n-2^n \cdot 7 n-2^n $

$ \quad=\left(2^n+{ }^n C_1 2^{n-1} \cdot 14+{ }^n C_2 2^{n-2} \cdot 14^2+\ldots+14^n\right)-2^n \cdot 7 n-2^n $

$ \quad=14^2\left({ }^n C_2 2^{n-2}+{ }^n C_3 2^{n-3} 14+\ldots+14^{n-2}\right)+\left(2^n+{ }^n C_1 \cdot 2^{n-1} \cdot 14-2^n \cdot 7 n-2^n\right) $

$ \quad=14^2\left({ }^n C_2 2^{n-2}+{ }^n C_3 2^{n-3} \cdot 14+\ldots+14^{n-2}\right)+\left(2^n+n 2^{n-1} \cdot 2^1 \cdot 7-2^n \cdot 7 n-2^n\right) $

$ \quad=14^2\left({ }^n C_2 \cdot 2^{n-2}+{ }^n C_3 \cdot 2^{n-3} \cdot 14+\ldots+14^{n-2}\right) $

This is divisible by $14^2$

Hence, the answer is the option 3.

Example 2: The number $101^{100}-1$ is divisible by

1) 101

2) 1001

3) 10000

4) 100000

Solution:

$ (101)^{100}-1 $

$=(1+100)^{100}-1$

$ =\left({ }^{100} \mathrm{C}_0+{ }^{100} \mathrm{C}_1(100)+\ldots\right)-1 $

$ =10000\left(1+{ }^{100} \mathrm{C}_2+\ldots\right) $

obviously 10000 is the correct choice.

Hence, the answer is the option (3).

Example 3: $(1+x)^n-n x-1$ is divisible by (where $n \in N$ ):

1) $2 x$

2) $x^2$

3) $2 x^3$

4) All of these

Solution:

$(1+x)^n=1+n x+\frac{n(n-1)}{2!} x^2+\frac{n(n-1)(n-2)}{3!} x^3+\ldots $

$ \Rightarrow(1+x)^n-n x-1 $

$ =x^2\left[\frac{n(n-1)}{2!}+\frac{n(n-1)(n-3)}{3!} x+\ldots .\right] $

From above it is clear that $(1+x)^n-n x-1$ is divisible by $x^2$.

Trick: $(1+x)^n-n x-1$. Put $\mathrm{n}=2$ and $\mathrm{x}=3$; Then $4^2-2.3-1=9$ is not divisible by 6,54 but divisible by 9 . Which is given by option (b) i.e., $x^2=9$

Hence, the answer is the option (2).

Example 4: $49^n+16 n-1$ is divisible by:

1) 3

2) 19

3) 64

4) 29

Solution:

$49^n+16 n-1=(1+48)^n+16 n-1 $

$ 1+{ }^n C_1(48)+{ }^n C_2(48)^2+\ldots+{ }^n C_n(48)^n+16 n-1$

$= (48 n+16 n)+{ }^n C_2(48)^2+{ }^n C_3(48)^3+\ldots+{ }^n C_n(48)^n $

$= 64 n+8^2\left[{ }^n C_2 \cdot 6^2+{ }^n C_3 \cdot 6^3 \cdot 8+{ }^n C_4 \cdot 6^4 \cdot 8^2+\ldots+{ }^n C_n \cdot 6^n \cdot 8^{n-2}\right] $

Hence, $49^n+16 n-1$ is divisible by 64 .

Hence, the answer is option (3).

Example 5: The greatest integer $m$ such that $5^m$ divides $7^{2 n}+2^{3 n-3} \cdot 3^{n-1}$ for $n \in N$, is

1) 0

2) 1

3) 2

4) None of these

Solution:

$ 7^{2 n}+2^{3 n-3} \cdot 3^{n-1}=49^n+24^{n-1} $

$ =(50-1)^n+(25-1)^{n-1} $

$ =\text { Multiple of } 50+(-1)^n+\text { Multiple of } 25+(-1)^{n-1}$

$ =\text { a multiple of } 25, \text { since }(-1)^n+(-1)^{n-1}=0 $

Hence, the answer is option (1).

Frequently Asked Questions (FAQs)

1. What is the divisibility rule for binomial expansion?

The divisibility rule for binomial expansion states that in the expansion of (a + b)^n, where n is a positive integer, each term is divisible by both a and b if and only if n is even. This rule helps us understand the structure and properties of binomial expansions without having to calculate every term.

2. Why does the divisibility rule only apply when the exponent is even?

The divisibility rule applies only when the exponent is even because of the symmetry in the expansion. When n is even, each term in the expansion has an even number of factors, ensuring that both a and b appear an even number of times in each term, making them divisible by both a and b.

3. How does the divisibility rule relate to the symmetry of Pascal's triangle?

The divisibility rule is closely related to the symmetry of Pascal's triangle. In Pascal's triangle, even-numbered rows (starting from row 0) have coefficients that are all even, except for the first and last terms. This symmetry corresponds to the divisibility of terms in binomial expansions with even exponents.

4. Can you explain why (a + b)^n is always divisible by (a + b)?

(a + b)^n is always divisible by (a + b) because it can be factored as (a + b)[(a + b)^(n-1)]. This is true for any positive integer n, not just even ones. The divisibility by (a + b) is a fundamental property of binomial expansions and is independent of the divisibility rule for individual terms.

5. How does the divisibility rule affect the factorization of binomial expansions?

The divisibility rule affects factorization by revealing common factors in the expansion. When n is even, we can factor out (ab) from the entire expansion, as every term is divisible by both a and b. This simplifies the expression and helps in identifying the structure of the expansion.

6. What is the relationship between the divisibility rule and the binomial theorem?

The divisibility rule is a consequence of the binomial theorem. The binomial theorem provides the general form for expanding (a + b)^n, and the divisibility rule is a property that emerges from this expansion when n is even. Understanding the divisibility rule deepens our comprehension of the binomial theorem's implications.

7. How can the divisibility rule be used to check for errors in binomial expansions?

The divisibility rule can be used as a quick check for errors in binomial expansions with even exponents. If any term in the expansion is not divisible by both a and b when n is even, it indicates an error in the calculation or expansion process. This serves as a useful verification tool.

8. What happens to the divisibility of terms when the exponent is odd?

When the exponent is odd, the terms in the binomial expansion are not all divisible by both a and b. Instead, they alternate between being divisible by a and b. This alternating pattern is due to the odd number of factors in each term when n is odd.

9. How does the divisibility rule relate to the concept of even and odd functions?

The divisibility rule relates to even and odd functions in that binomial expansions with even exponents behave like even functions (symmetric about the y-axis when graphed), while those with odd exponents behave like odd functions (symmetric about the origin). This connection helps in understanding the behavior of binomial expressions graphically.

10. Can the divisibility rule be extended to multinomial expansions?

The concept of the divisibility rule can be extended to multinomial expansions, but it becomes more complex. In general, for a multinomial expansion (a + b + c + ...)^n, if n is divisible by the number of terms in the base expression, then each term in the expansion will be divisible by the product of all base terms. However, this is a more advanced concept and requires careful consideration of the specific multinomial.

11. How does the divisibility rule affect the remainder when dividing a binomial expansion by (a + b)?

The divisibility rule doesn't directly affect the remainder when dividing by (a + b), as all binomial expansions (a + b)^n are always divisible by (a + b) without a remainder. However, understanding the divisibility rule can help in simplifying the quotient, especially when n is even, as it reveals additional common factors.

12. What is the connection between the divisibility rule and the concept of perfect squares in algebra?

The divisibility rule is closely connected to perfect squares in algebra. When n = 2, the binomial expansion (a + b)^2 = a^2 + 2ab + b^2 is a perfect square, and each term is divisible by both a and b. This is the simplest case of the divisibility rule and forms the basis for understanding more complex even-powered expansions.

13. How can the divisibility rule help in solving problems involving binomial probabilities?

In binomial probability problems, the divisibility rule can help simplify calculations, especially when dealing with even numbers of trials. It allows for easier factorization of probability expressions, which can lead to more efficient problem-solving and a clearer understanding of the probability distribution's structure.

14. Why doesn't the divisibility rule apply to fractional or negative exponents in binomial expansions?

The divisibility rule doesn't apply to fractional or negative exponents because these expansions result in infinite series rather than finite polynomials. The rule is based on the properties of integer exponents and the finite number of terms in such expansions. Fractional and negative exponents produce different types of expressions that don't follow the same divisibility patterns.

15. How does the divisibility rule relate to the concept of polynomial long division?

The divisibility rule can be useful in polynomial long division involving binomial expressions. When dividing (a + b)^n by (a + b), if n is even, the divisibility rule ensures that each term in the quotient will be divisible by both a and b. This can simplify the division process and help in identifying patterns in the quotient.

16. Can you explain how the divisibility rule affects the coefficients in a binomial expansion?

The divisibility rule doesn't directly affect the coefficients in a binomial expansion, as these are determined by the binomial theorem. However, when n is even, the rule implies that each term (including its coefficient) must be divisible by ab. This can be used to check the validity of calculated coefficients and understand their properties.

17. How does the divisibility rule help in understanding the behavior of binomial expansions at extreme values?

The divisibility rule helps in understanding the behavior of binomial expansions at extreme values by revealing the structure of the terms. For even n, since all terms are divisible by both a and b, the expansion will approach zero if either a or b approaches zero. This insight is valuable in analyzing the limits and asymptotic behavior of binomial expressions.

18. What is the relationship between the divisibility rule and the concept of algebraic identities?

The divisibility rule is closely related to algebraic identities, particularly those involving even powers. For example, the identity (a^2 - b^2) = (a+b)(a-b) is a consequence of the divisibility rule for n=2. Understanding the rule helps in deriving and applying more complex algebraic identities involving binomial expressions.

19. How can the divisibility rule be used to simplify complex algebraic expressions involving binomials?

The divisibility rule can be used to simplify complex algebraic expressions by identifying common factors in binomial expansions with even exponents. By recognizing that all terms are divisible by both a and b when n is even, we can factor out (ab) from the entire expression, often leading to more manageable forms and revealing underlying patterns.

20. What role does the divisibility rule play in understanding the graphical representation of binomial functions?

The divisibility rule plays a role in understanding the graphical representation of binomial functions by revealing symmetry properties. For even n, the rule implies that the graph will have even symmetry (symmetric about the y-axis), while for odd n, this symmetry is not present. This helps in sketching and analyzing the behavior of binomial functions visually.

21. How does the divisibility rule relate to the concept of polynomial roots?

The divisibility rule relates to polynomial roots in that it provides insights into the factors of binomial expansions. For even n, since all terms are divisible by both a and b, we know that -a/b is always a root of the equation (a + b)^n = 0. This connection helps in understanding the nature and distribution of roots in binomial equations.

22. Can you explain how the divisibility rule affects the process of polynomial factorization?

The divisibility rule affects polynomial factorization by revealing common factors in binomial expansions with even exponents. When n is even, we can always factor out (ab) from the entire expansion. This simplifies the factorization process and helps in identifying the structure of the polynomial, often leading to more efficient factorization strategies.

23. How does the divisibility rule help in understanding the concept of algebraic degree in binomial expansions?

The divisibility rule helps in understanding algebraic degree by highlighting the relationship between the exponent n and the terms in the expansion. For even n, all terms have factors of both a and b, which means the lowest degree term is (ab)^(n/2). This insight helps in analyzing the degree and structure of binomial expansions without full calculation.

24. What is the significance of the divisibility rule in combinatorial problems involving binomial coefficients?

In combinatorial problems, the divisibility rule reflects properties of binomial coefficients. For even n, the rule implies that the product of any binomial coefficient and (ab)^(n/2) is divisible by (ab)^n. This connection helps in solving complex counting problems and understanding the properties of combinations and Pascal's triangle.

25. How can the divisibility rule be used to predict patterns in the expansion of (a + b)^n for large values of n?

The divisibility rule helps predict patterns for large n by ensuring that when n is even, all terms will be divisible by (ab)^(n/2). This allows us to factor out this common term and focus on the remaining pattern of coefficients and powers, which follow Pascal's triangle. It simplifies the analysis of the expansion's structure for large exponents.

26. What is the relationship between the divisibility rule and the concept of polynomial interpolation?

The divisibility rule relates to polynomial interpolation by providing insights into the structure of binomial expansions. When interpolating points from a binomial function with an even exponent, the rule suggests that the resulting polynomial will have certain divisibility properties. This can guide the choice of interpolation points and help verify the accuracy of the interpolated polynomial.

27. How does the divisibility rule affect the process of solving binomial equations?

The divisibility rule affects solving binomial equations by revealing structural properties. For equations of the form (a + b)^n = k where n is even, the rule ensures that k must be divisible by (ab)^(n/2) for real solutions to exist. This insight can simplify the solving process and help in determining the nature of solutions without extensive calculation.

28. Can you explain how the divisibility rule relates to the concept of polynomial identities?

The divisibility rule is closely related to polynomial identities, particularly those involving even powers of binomials. For example, the identity (a + b)^4 - (a - b)^4 = 8ab(a^2 + b^2) is a consequence of the divisibility rule for n=4. Understanding the rule helps in deriving and verifying more complex polynomial identities involving binomial expressions.

29. How does the divisibility rule help in understanding the behavior of binomial expansions under different scaling factors?

The divisibility rule helps understand scaling behavior by showing how terms change when a and b are scaled. For even n, if we scale a and b by a factor k, the entire expansion is scaled by k^n, and each term remains divisible by both scaled variables. This insight is useful in analyzing how binomial expressions respond to scaling in various applications.

30. What is the connection between the divisibility rule and the concept of polynomial decomposition?

The divisibility rule is connected to polynomial decomposition by revealing factors common to all terms in even-powered binomial expansions. This allows for a natural decomposition of the polynomial into a product of (ab)^(n/2) and a remaining polynomial. Understanding this connection helps in more advanced techniques of polynomial decomposition and analysis.

31. How can the divisibility rule be applied in the study of algebraic structures like rings and fields?

In the study of algebraic structures, the divisibility rule provides insights into the behavior of binomial elements in rings and fields. For even n, it ensures certain divisibility properties that can be used to analyze ideals, factor rings, and field extensions generated by binomial elements. This application extends the rule's relevance to more abstract algebraic contexts.

32. What role does the divisibility rule play in understanding the convergence of binomial series?

While the divisibility rule primarily applies to finite expansions, it provides insights into the behavior of infinite binomial series. For series with even-powered terms, the rule suggests patterns in the divisibility of terms, which can be useful in analyzing convergence properties, especially when the series involves rational functions of a and b.

33. How does the divisibility rule relate to the concept of algebraic invariants in binomial expansions?

The divisibility rule relates to algebraic invariants by identifying properties that remain unchanged under certain transformations. For even n, the divisibility of all terms by both a and b is an invariant property of the expansion. This concept of invariance is crucial in more advanced algebraic studies and can lead to deeper insights into the structure of binomial expressions.

34. Can you explain how the divisibility rule affects the analysis of binomial probability distributions?

In binomial probability distributions, the divisibility rule affects the analysis of even-numbered trials. It ensures that probability expressions for even n will have certain divisibility properties, which can simplify calculations and reveal symmetries in the distribution. This can be particularly useful in analyzing the central tendencies and spread of binomial distributions.

35. How does the divisibility rule help in understanding the concept of algebraic closure in relation to binomial equations?

The divisibility rule provides insights into the nature of roots of binomial equations, which relates to algebraic closure. For even n, the rule ensures that certain types of roots always exist (e.g., -a/b), which helps in understanding what additional elements might be needed to achieve algebraic closure when working with binomial expressions in different number systems.

36. What is the significance of the divisibility rule in the study of symmetric polynomials?

The divisibility rule is significant in the study of symmetric polynomials as it reveals symmetry properties in even-powered binomial expansions. These symmetries are related to the invariance of the expansion under the interchange of a and b, which is a key property of symmetric polynomials. This connection helps in analyzing more complex symmetric polynomial structures.

37. How can the divisibility rule be used to simplify calculations in modular arithmetic involving binomials?

In modular arithmetic, the divisibility rule can simplify calculations with binomial expansions. For even n, knowing that all terms are divisible by both a and b mod m (where m is the modulus) can reduce the complexity of computations. This is particularly useful in cryptography and coding theory where modular binomial expressions are common.

38. What is the relationship between the divisibility rule and the concept of polynomial congruences?

The divisibility rule relates to polynomial congruences by providing information about the divisibility properties of binomial expansions in modular arithmetic. For even n, it ensures certain congruence properties that can be exploited to solve polynomial congruences involving binomials more efficiently, especially in number theory applications.

39. How does the divisibility rule contribute to our understanding of the algebraic structure of binomial expansions?

The divisibility rule contributes to our understanding of the algebraic structure of binomial expansions by revealing inherent patterns and properties. It shows that for even n, the expansion has a multiplicative structure involving (ab)^(n/2), which helps in analyzing the overall algebraic form and behavior of the expansion without explicit calculation.

40. Can you explain how the divisibility rule relates to the concept of polynomial interpolation over finite fields?

In polynomial interpolation over finite fields, the divisibility rule provides insights into the behavior of binomial functions. For even n, it ensures certain divisibility properties in the field, which can guide the choice of interpolation points and help in verifying the correctness of the interpolated polynomial, especially when working with binomial-based functions in finite field applications.

41. How does the divisibility rule help in analyzing the factorization patterns of binomial expansions over different number systems?

The divisibility rule helps in analyzing factorization patterns by revealing common factors in even-powered expansions across different number systems. Whether working in real, complex, or p-adic number systems, the rule provides a consistent starting point for factorization, helping to identify system-specific factorization behaviors and patterns.

42. What is the connection between the divisibility rule and the concept of algebraic independence in binomial expressions?

The divisibility rule relates to algebraic independence by showing that for even n, a and b are not algebraically independent in the expansion of (a + b)^n, as every term is