Results on Binomial Theorem of any Index

An expression with two terms is called the binomial expansion. In the case of higher degree expression, it is difficult to calculate it manually. In these cases, Binomial theorem can be used to calculate it. Binomial theorem is used for the expansion of a binomial expression with a higher degree. Binomial coefficients are the coefficients of the terms in the Binomial expansion. Binomial theorem is proved using the concept of mathematical induction. Apart from Mathematics, Binomial theorem is also used in statistical and financial data analysis.

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This article is about the binomial inside binomial which falls under the broader category of Binomial Theorem and its applications. It is one of the important topics for competitive exams.

Binomial Theorem for any index

Statement: If $n$ is a rational number and $x$ is a real number such that $|\mathrm{x}|<1$, then,

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

Proof:

Let $f(x)=(1+x{)}^n=a_0+a_{1}x+a_2{x}^2+\dots +a_1{x}^n+\dots$

$f(0)=(1+0n=1$

Differentiating (1) w.r.t. $x$ on both sides, we get

$=a_1+2a_{2}x+3a_3{x}^3+4a_4{x}^3+\dots +r{a}_1{x}^r-1+\dots (2)$

Put $x=0$, we get $n=a_1$

Differentiating (2) w.r.t. $\times$ on both sides, we get

$=2a_2+6a_{3}x+12a_4{x}^2+\dots +r(r-1)a_4{x}^h-2+\dots$

Put $x=0$, we get $a_2=[n(n-1)]/2$ !

Differentiating (3), w.r.t. x on both sides, we get

Put $x=0$, we get $a_3=[n(n-1)(n-2)]/3$ !

Similarly, we get $a_4=[n(n-1)(n-2)(n-3)]/r!$ and so on

$\therefore a_n=[n(n-1)(n-2)\dots (n-r+1)]/r!$

Putting the values of $a_0,a_1,a_2,a_3,\dots ,a_n$ obtained in (1), we get

$(1+xn=1+nx+[\{n(n-1)\}/2!]x^2+[\{n(n-1)(n-2)\}/2!]x^3+\dots +[\{n(n-1)(n-2)\dots (n-r+$

1) $\}/r!]x^{[}+\dots$

Hence proved the Binomial theorem of any index.

Results on Binomial Theorem of any Index

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

In the above expansion replace ' $n$ ' with ' $-n$ '

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

$\dots +\frac{(-n)((-n)-1)((-n)-2)\dots ((-n)-r+1)}{r!}{x}^r\dots \dots \mathrm{\infty }$

$\Rightarrow (1+\mathrm{x}{)}^{-\mathrm{n}}=1-\mathrm{nx}+\frac{\mathrm{n}(\mathrm{n}+1)}{2!}{\mathrm{x}}^2-\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{3!}{\mathrm{x}}^3+\dots \dots$

$\dots +(-1{)}^r\frac{n(n+1)(n+2)\dots .(n+r-1)}{r!}{x}^r\dots \dots .\mathrm{\infty }$

If $-n$ is a negative integer (so that $n$ is a positive integer), then we can re-write this expression as

$=1-{}^n{C}_{1}x+{}^{n+1}{C}_2{x}^2-{}^{n+2}{C}_3{x}^3+\cdots +{}^{n+r-1}{C}_r(-x{)}^r+\cdots$

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

$+\frac{n(n+1)(n+2)\cdots (n+r-1)}{r!}{x}^r+\cdots$

If $-n$ is a negative integer (so that $n$ is a positive integer), then we can re-write this expression as

$=1+{}^n{C}_{1}x+{}^{n+1}{C}_2{x}^2+{}^{n+2}{C}_3{x}^3+\cdots +{}^{n+r-1}{C}_r(x{)}^r+\cdots$

Important Note:

The coefficient of $x^5$ in $(1-x{)}^{-n}$, (when $n$ is a natural number) is ${}^{n+r-1}{C}_r$

Some Important Binomial Expansion

1. $(1+x{)}^{-1}=1-x+x^2-x^3+\cdots$

2. $(1-x{)}^{-1}=1+x+x^2+x^3+\cdots$

3. $(1+x{)}^{-2}=1-2x+3x^2-4x^3+\cdots$

4. $(1-x{)}^{-2}=1+2x+3x^2+4x^3+\cdots$

Summary

Binomial theorem is used for the expansion of a binomial expression with a higher degree. Binomial coefficients are the coefficients of the terms in the Binomial expansion. Binomial coefficients of the term equidistant from the beginning and end are equal. Understanding the product of two binomial coefficients gives an idea to solve more complex problems not only in calculus, statistics, data analysis etc.

Recommended Video Based on Results on Binomial Theorem of any Index:

Solved Examples based on Results on Binomial Theorem of any Index

Example 1: If the expansion in powers of $x$ of the function $\frac{1}{(1-ax)(1-bx)}$ is $a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\dots \dots$. , then $a_n$ is

1) $\frac{b^n-a^n}{b-a}$

2) $\frac{a^n-b^n}{b-a}$

3) $\frac{a^{n+1}-b^{n+1}}{b-a}$

4) $\frac{b^{n+1}-a^{n+1}}{b-a}$

Solution:

As we learned,

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

And

$(1+x{)}^{-1}=1-x+x^2-x^3+\cdots$

Now,

Expansion of $(1-ax{)}^{-1}(1-bx{)}^{-1}$

$\Rightarrow (1+ax+a^2{x}^2+a^3{x}^3+\dots \dots \dots +a^n{x}^n+\dots \dots )(1+bx+b^2{x}^2+\dots \dots \dots +b^n{x}^n+\dots \dots )$

Coefficient of $x_n\Rightarrow a^n+a^{n-1}b+a^{n-2}{b}^2+\dots \dots \dots +b^n$

$\Rightarrow \frac{b^{n+1}-a^{n+1}}{b-a}$ (Using sum of GP formula)

Hence, the answer is option (4).

Example 2: Let $[x]$ denote greatest integer less than or equal to x . If for $n\in N$, ${(1-x+x^3)}^n=\sum _{j=0}^{3n}{a}_j{x}^j,\sum _{\text{then }}^{\left[\frac{3n}{2}\right]}{a}_{2j}+4\sum _{j=0}^{\left[\frac{3n-1}{2}\right]}{a}_{2j+1}$ is equal to:

1) $2$

2) $n$

3) $1$

4) $2^{n-1}$

Solution:

${(1-x+x^3)}^n=\sum _{j=0}^{3n}{a}_j{x}^j$

${(1-x+x^3)}^n=a_0+a_{1}x+a_2{x}^2\dots \dots +a_{3n}{x}^{3n}$

$\sum _{j=0}^{\left[\frac{3n}{2}\right]}{a}_{2j}=\text{ Sum of }{a}_0+a_2+a_4\dots \dots .$

$\left[\frac{3n-1}{2}\right]$

$\sum _{j=0}{a}_{2j+1}=\text{ Sum of }{a}_1+a_3+a_5\dots \dots$

put $x=1$

$1=a_0+a_1+a_2+a_3\dots \dots \dots +a_{3a}$

Put $\mathrm{x}=-1$

$1=a_0-a_1+a_2-a_3\dots \dots \dots +(-1{)}^{3n}{a}_{3n}$

Solving (A) and (B)

$a_0+a_2+a_4\dots \dots =1$

$a_1+a_3+a_5\dots \dots =0$

$\sum _{j=0}^{\left[\frac{a}{2}\right]}{a}_{2j}+4\sum _{j=0}^{\left[\frac{s_n-1}{2}\right]}{a}_{2j+1}=1$

Hence, the answer is option 1.

Example 3: If $x$ is so small that $x^3$ and higher powers of $x$ may be neglected, then $\frac{(1+x{)}^{3/2}-{(1+\frac{1}{2}x)}^3}{(1-x{)}^{1/2}}$ may be approximated as:

1) $3x+\frac{3}{8}{x}^2$

2) $1-\frac{3}{8}{x}^2$

3) $\frac{x}{2}-\frac{3}{8}{x}^2$

4) $-\frac{3}{8}{x}^2$

Solution:

As we learned,

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

Now,

$\frac{((1+x{)}^{3/2}-{(1+\frac{x}{2})}^3)}{(1-x{)}^{1/2}}$

$\Rightarrow \frac{1+\frac{3x}{2}+\frac{3}{2}\cdot \frac{1}{2}\cdot \frac{x^2}{2}-1-\frac{3x}{2}-\frac{3\times 2}{2!}\frac{x^2}{4}}{(1-x{)}^{1/2}}$

$=(\frac{3x^2}{8}-\frac{3x^2}{4})(1-x{)}^{-1/2}=\frac{-3x^2}{8}(1-x{)}^{-1/2}$

$=\frac{-3x^2}{8}(1+\frac{x}{2}+\dots )$

$=\frac{-3x^2}{8}+\frac{-3x^3}{16}+\dots$

So, given expression $=\frac{-3x^2}{8}$ (as powers higher than 2 are neglected)

Hence, the answer is option (4).

Example 4: If $0<x<1$, then the first negative term in the expansion of $(1+x{)}^{41/7}$ is:

1) $5^{\circ }$ term

2) $8^n$ term

3) $6^{\text{n }}$ term

4) $7^{\ast }$ term

Solution:

$(1+x{)}^{\frac{11}{r}}$

$T_{r+1}=\frac{n(n-1)(n-2)\dots \dots \dots (n-r+1)}{r!}{x}^r$

For the first negative term, $(n-r+1)<0$

$n=\frac{41}{7}$

$r>6.85$

so, if $\mathrm{r}=7$

Then it would be the 8 th term

Hence, the answer is option (2).

Example 5: If the expansion in the power of $x$ of the function $\frac{1}{(1-ax)(1-bx)}$ is $a_0+a_{1}x+a_2{x}^2+\dots$, then $a_3$ is

1) ${}^{a_n}=\frac{a^n-b^n}{a-b}$

2) $a_n=\frac{a^{n+1}-b^{n+1}}{a-b}$

3) $a_n=\frac{a^{n-1}-b^{n-1}}{a-b}$

4) $1$

Solution:

Now

$(1-ax{)}^{-1}(1-bx{)}^{-1}=(1+ax+a^2{x}^2+\dots )(1+bx+b^2{x}^2+\dots )$

In the equation $a_0+a_{1}x+a_2{x}^2+\dots ,{\mathrm{a}}_n$ is the coefficient of $x^n$.

Coefficient of $x^n=a^n+a^{n-1}b+\dots +a{b}^{n-1}+b^n$

$a^n+a^{n-1}b+\dots a{b}^{n-1}+b^n$ is a GP with a common ratio $\frac{b}{a}$ and $(\mathrm{n}+1)$ terms Sum of the series is

$\frac{a^n[{\left(\frac{b}{a}\right)}^{n+1}-1]}{\frac{b}{a}-1}$

$\frac{a^n\frac{b^{n+1}-a^{n+1}]}{a^{n+1}}}{\frac{b-a}{a}}$

$a_n=\frac{a^{n+1}-b^{n+1}}{a-b}$

Hence, the answer is option 2.