Nature of Integral Part of Expression N = (a + sqrt b)^n

Binomial Theorem: Definition, Concepts, and Application in Integral Expressions

The Binomial Theorem is one of the most important algebraic resources in Class 11 Mathematics, widely used in JEE Main & Advanced, CUET, NDA, and other competitive exams. It provides a systematic method to expand expressions of the form $(a+b)^n$ without repeated multiplication and forms the foundation for concepts like binomial coefficients, general term, and greatest term.

In addition to standard expansions, special expressions such as $(a+\sqrt{b})^n$ and $(a-\sqrt{b})^n$ play a crucial role in problems involving the integral part and fractional part of algebraic expressions. These identities are frequently tested in higher-order binomial theorem questions, especially in integer-based and maximum–minimum problems.

Below is a structured explanation covering the Binomial Theorem, properties of binomial coefficients, and the method to find the integral part of expressions involving square roots, aligned with exam requirements and quick revision needs.

Understanding the Binomial Theorem

When you have an expression like $(a + b)^n$, expanding it directly becomes extremely lengthy for large values of $n$. The Binomial Theorem gives a shortcut.

General Expansion Formula

For any positive integer $n$:

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

Each term follows a consistent pattern:

• Powers of $a$ decrease,

• Powers of $b$ increase,

• Coefficients come from binomial coefficients.

Binomial Coefficients and Their Properties

The binomial coefficient $\binom{n}{r}$ or $^nC_r$ represents the number of ways to choose $r$ elements from $n$.
It is calculated using:

$\binom{n}{r} = \frac{n!}{r!(n-r)!}$

Key Properties Useful for Exams

• $\binom{n}{r} = \binom{n}{n-r}$

• $\binom{n}{0} = \binom{n}{n} = 1$

• All coefficients in an expansion are positive integers.

• They form Pascal’s Triangle.

These properties help simplify expressions, find specific terms, and solve combinatorics problems efficiently.

Integral Part Problems: Expressions of the Form $(a + \sqrt{b})^n$

One of the most interesting applications of the binomial theorem is determining the integral part of expressions like:

$N = (a + \sqrt{b})^n$

Even though the expression includes a square root and looks non-integer, the integral part can be found using conjugates.

Why These Expressions Are Special

For numbers such as $(a + \sqrt{b})$ where $a$ and $b$ are integers but $\sqrt{b}$ is irrational, raising them to a power still creates structured patterns.

A key idea:

• The irrational part cancels out when we consider both $(a + \sqrt{b})^n$ and $(a - \sqrt{b})^n$ together.

This is why these problems often appear in JEE and Olympiad-level algebra.

Key Properties of $(a+\sqrt{b})^n$ and $(a-\sqrt{b})^n$

Expressions of the form $(a+\sqrt{b})^n$ and $(a-\sqrt{b})^n$ are called conjugate binomial expressions. They are extremely important in Class 12 Binomial Theorem, especially for questions involving the integral part, fractional part, and integer-valued expressions. These properties are frequently used in JEE, CUET, NDA, and board-level exams.

Why $(a-\sqrt{b})^n$ Lies Between 0 and 1

If $a$ and $b$ are positive real numbers such that $a^2 - b = 1$

Then, $(a+\sqrt{b})(a-\sqrt{b}) = 1$

This implies, $a-\sqrt{b} = \dfrac{1}{a+\sqrt{b}}$

Since $a+\sqrt{b} > 1$, we get $0 < a-\sqrt{b} < 1$

Raising both sides to a positive integer power $n$, $0 < (a-\sqrt{b})^n < 1$

This result is crucial because it guarantees that $(a-\sqrt{b})^n$ contributes only a fractional value, which plays a key role in identifying the integral part of $(a+\sqrt{b})^n$.

Sum of Conjugate Expressions and Integer Value Property

Using the Binomial Theorem, $(a+\sqrt{b})^n + (a-\sqrt{b})^n$

On expansion, all odd powers of $\sqrt{b}$ cancel out, leaving only even-powered terms.

Thus, the sum becomes: $(a+\sqrt{b})^n + (a-\sqrt{b})^n = 2\left[{}^nC_0 a^n + {}^nC_2 a^{n-2} b + {}^nC_4 a^{n-4} b^2 + \cdots \right]$

If $a$ and $b$ are integers, then the right-hand side is an integer.

This property is widely used to show that: $(a+\sqrt{b})^n + (a-\sqrt{b})^n$ is an integer and helps determine the greatest integer less than $(a+\sqrt{b})^n$.

Product of Conjugate Expressions and Its Applications

The product of conjugate expressions is given by: $(a+\sqrt{b})^n (a-\sqrt{b})^n = (a^2 - b)^n$

If $a^2 - b = 1$, then $(a+\sqrt{b})^n (a-\sqrt{b})^n = 1$

This identity is extremely powerful in integral part and fractional part problems, allowing us to write: $(a-\sqrt{b})^n = \dfrac{1}{(a+\sqrt{b})^n}$

This directly leads to results such as:

If $(a+\sqrt{b})^n = I + f$ where $0<f<1$, then $(a-\sqrt{b})^n = 1 - f$ and hence, $(a+\sqrt{b})^n (1 - f) = 1$

Step-by-Step Method to Find the Integral Part

This section walks you through an easy, systematic method to find the integral part of expressions like $(a + \sqrt{b})^n$. By using conjugates and basic algebraic properties, you can determine the integer value without expanding the entire expression.

Step 1: Choose the appropriate conjugate

Define the conjugate expression as:

• $N' = (a - \sqrt{b})^n$ if $a > \sqrt{b}$

• $N' = (\sqrt{b} - a)^n$ if $\sqrt{b} > a$

This helps eliminate $\sqrt{b}$ in the next step.

Step 2: Form an integer by adding or subtracting

Both expressions:

$(a + \sqrt{b})^n + (a - \sqrt{b})^n$
$(a + \sqrt{b})^n - (a - \sqrt{b})^n$

always produce integers because the irrational parts cancel out.

Thus, you obtain a clean integer relation involving $N$.

Step 3: Write the number in standard form

Write:

$N = I + f$

where:

• $I = [N]$ is the integer part

• $f = {N}$ is the fractional part

Then use the integer expression from Step 2 to solve for $I$.

Example: Integral Part of $(3\sqrt{3} + 5)^{2n+1}$

Let:

$P = (3\sqrt{3} + 5)^{2n+1}$

Step 1: Conjugate Expression

$P' = (3\sqrt{3} - 5)^{2n+1}$

Since $3\sqrt{3} - 5$ is a number between 0 and 1,
we have $0 < P' < 1$.

Step 2: Subtract to remove irrational components

Consider:

$P - P' = 2\left[\binom{2n+1}{1}(3\sqrt{3})^{2n}5 + \binom{2n+1}{3}(3\sqrt{3})^{2n-2}5^3 + \ldots\right]$

This entire right side is an even integer.
Let that be $2k$.

Step 3: Use fractional part properties

Write $P = I + f$.

Then:

$I + f - P' = 2k$

Since both $f$ and $P'$ lie between 0 and 1:

$-1 < f - P' < 1$

But $f - P'$ must also be an integer.
The only integer between −1 and 1 is 0.

So:

$f - P' = 0$
$I = 2k$

This shows that the integer part is always even.

Final Result

The integral part of $(3\sqrt{3} + 5)^{2n+1}$ is always an even integer, regardless of the value of $n$.

Solved Examples based on Nature of Integral Part of Expression $N = (a + \sqrt {b})^n$

Example 1:
If $P=(7+4\sqrt{3})^9$ and $P=I+f$, where $I$ is the integer just less than $P$, then the value of $P(1-f)$ is

1. $1$

2. $2^n$

3. $2$

4. $2^{2n}$

Solution

$P=(7+4\sqrt{3})^9$

Let $P'=(7-4\sqrt{3})^9$

Since $0<7-4\sqrt{3}<1 \Rightarrow 0<P'<1$

Using binomial expansion,

$P+P' = 2\left[{}^9C_0 7^9 + {}^9C_2 7^7(4\sqrt{3})^2 + {}^9C_4 7^5(4\sqrt{3})^4 + \cdots \right]$

Hence, $P+P'$ is an even integer.

Given $P=I+f$ where $0<f<1$

$I+f+P'$ is an even integer.

Since $I$ is an integer,
$f+P'$ must be an integer.

Now, $0<f<1$ and $0<P'<1$

So, $0<f+P'<2$

The only integer in this interval is $1$.

Hence, $f+P'=1$

$P'=1-f$

Now, $P(1-f)=PP'$

$P(1-f)=(7+4\sqrt{3})^9(7-4\sqrt{3})^9$

$=[(7+4\sqrt{3})(7-4\sqrt{3})]^9$

$=(49-48)^9$

$=1^9=1$

Hence, the correct answer is option (1).

Example 2:
If $P=(10+3\sqrt{11})^n$ and $P'=(10-3\sqrt{11})^n$, then which of the following is not true?

1. $PP'=1$

2. $P+P'$ is an even integer

3. $P'=1-f$, where $f$ is the fractional part of $P$

4. $f+P' \in (0,1)$

Solution

i) $PP'=(10+3\sqrt{11})^n(10-3\sqrt{11})^n$

$=(10^2-(3\sqrt{11})^2)^n$

$=(100-99)^n$

$=1$

So, statement (1) is true.

ii) $P+P' = 2\left[{}^nC_0 10^n + {}^nC_2 10^{n-2}(3\sqrt{11})^2 + {}^nC_4 10^{n-4}(3\sqrt{11})^4 + \cdots \right]$

Hence, $P+P'$ is an even integer.

So, statement (2) is true.

iii) Let $P=[P]+f$, where $[P]$ is the greatest integer less than $P$

Since $P+P'$ is an integer,

$[P]+f+P'$ is an integer

Thus, $f+P'$ is an integer.

Since $0<f<1$ and $0<P'<1$,

$f+P'=1$

So, $P'=1-f$

Statement (3) is true.

iv) $f+P'=1 \notin (0,1)$

Hence, statement (4) is not true.

Correct answer: option (4).

Example 3:
If $x=(7+4\sqrt{3})^{2n}=[x]+f$, then $x(1-f)$ is equal to

1. $1$

2. $2$

3. $3$

4. $0$

Solution

Since $(7-4\sqrt{3})=\dfrac{1}{7+4\sqrt{3}}$

We have $0<7-4\sqrt{3}<1$

Let $F=(7-4\sqrt{3})^{2n}$

Then, $0<F<1$

Now, $x+F=(7+4\sqrt{3})^{2n}+(7-4\sqrt{3})^{2n}$

$=2\left[{}^{2n}C_0 7^{2n} + {}^{2n}C_2 7^{2n-2}(4\sqrt{3})^2 + \cdots + {}^{2n}C_{2n}(4\sqrt{3})^{2n} \right]$

$=2m$, where $m$ is an integer

Thus, $[x]+f+F=2m$

$f+F=2m-[x]$

Since $0<f<1$ and $0<F<1$

$0<f+F<2$

Hence, $f+F=1$

Now, $x(1-f)=xF$

$=(7+4\sqrt{3})^{2n}(7-4\sqrt{3})^{2n}$

$=(49-48)^{2n}$

$=1$

Hence, the correct answer is option (1).

Example 5:
Let $R=(5\sqrt{5}+11)^{2n+1}$ and $f=R-[R]$, where $[ ]$ denotes the greatest integer function. Find $R\cdot f$.

1. $4^{2n+1}$

2. $4^{2n}$

3. $4^{2n-1}$

4. $4^{-2n}$

Solution

$R=[R]+f$

Let $f'=(5\sqrt{5}-11)^{2n+1}$

Since $0<5\sqrt{5}-11<1$

We have $0<f'<1$

Now, $f+[R]-f'=(5\sqrt{5}+11)^{2n+1}-(5\sqrt{5}-11)^{2n+1}$

$=2\left[{}^{2n+1}C_1(5\sqrt{5})^{2n}11 + {}^{2n+1}C_3(5\sqrt{5})^{2n-2}11^3 + \cdots \right]$

$=2K$, where $K$ is an integer

Thus, $f-f'=2K-[R]$

Since $-1<f-f'<1$

We get $f-f'=0$

So, $f=f'$

Now, $R\cdot f=(5\sqrt{5}+11)^{2n+1}(5\sqrt{5}-11)^{2n+1}$

$=(125-121)^{2n+1}$

$=4^{2n+1}$

Hence, the correct answer is option (1).

Last Digits and Remainder using the Binomial Expansion