# Binomial Theorem

Binomial Theorem – As the power increases the expansion becomes lengthy and tedious to calculate. A binomial expression that has been raised to a very large power can be easily calculated with the help of Binomial Theorem.

## Introduction to the Binomial Theorem

Binomial Expression: A binomial expression is an algebraic expression which contains two dissimilar terms. Ex: $a+b,a^{3}+b^{3}$ etc.

Binomial Theorem: Let $n\epsilon N,x,y,\epsilon R$ then

$(x+y)^{n}=\sum_{r=0}^{n}nC_{r}x^{n-r}.y^{r}+nC_{r}x^{n-r}y^{r}+…+nC_{n-1}x.y^{n-1}+nC_{n}.y^{n}$

i.e. $(x+y)^{n}=\sum_{r=0}^{n}nC_{r}x^{n-r}.y^{r}$ where

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

Illustration 1: Expand $(\frac{x}{3}+\frac{2}{y})^{4}$

Sol:

$(\frac{x}{3}+\frac{2}{y})^{4}$ = $4c_{_{o}}\left ( \frac{x}{3} \right )^{4}+4c_{_{1}}\left (\frac{x}{3}\right )^{3}\left (\frac{2}{y}\right )+4c_{_{2}}\left (\frac{x}{3}\right )^{2}\left (\frac{2}{y}\right)^{2}+4c_{_{3}}\left (\frac{x}{3}\right )\left (\frac{2}{y} \right)^{3}+4c_{_{4}}\left (\frac{2}{y} \right )^{4}$

⇒ $\frac{x^{4}}{81}+\frac{8x^{3}}{27y}+\frac{8x^{2}}{3y^{2}}+\frac{32x}{3y^{3}}+\frac{16}{y^{4}}$

Illustration 2: $\left ( \sqrt{2}+1 \right )^{5}+\left ( \sqrt{2}-1 \right )^{5}=$

Sol:

We have $(x+y)^{5}+(xy)^{5}=2[5C_{0}x^{5}+5C_{2}x^{3}y^{2}+5C_{4}xy^{4}]$

$=2(x^{5}+10x^{3}y^{2}+5xy^{4})$

Now $\left ( \sqrt{2}+1 \right )^{5}+\left ( \sqrt{2}-1 \right )^{5}=2[(\sqrt{2})^{5}+10(\sqrt{2})^{3}(1)^{2}+5(\sqrt{2})(1)^{4}]$

=$58\sqrt{2}$

## Properties of Binomial Expansion

• The total number of terms in the expansion of $(x+y)^{n}$ are $(n+1)$
• The sum of exponents of x and y is always n.
• $nC_{0},nC_{1},nC_{2},…nC_{n}$ are called binomial coefficients and also represented by $C_{0},C_{1},C_{2},…C_{n}$
• The binomial coefficients which are equidistant from the beginning and from the ending are equal i.e. $nC_{0}=nC_{n},nC_{1}=nC_{n-1},nC_{2}=nC_{n-2}…..$ Etc.

To find binomial coefficients we can use Pascal Triangle also.

Some other useful expansions:

• $(x+y)^{n}+(x-y)^{n}=2[C_{o}x^{n}+C_{2}x^{n-1}y^{2}+C_{4}x^{n-4}y^{4}+…]$
• $(x+y)^{n}-(x-y)^{n}=2[C_{1}x^{n-1}y+C_{3}x^{n-3}y^{3}+C_{5}x^{n-5}y^{5}+…]$
• $(1+x)^{n}=\sum_{r-0}^{n}nC_{r}.x^{r}=C_{o}+C_{1}x+C_{2}x^{2}+…C_{n}x^{n}]$
• $(1+x)^{n}+(1-x)^{n}=2[C_{o}+C_{2}x^{2}+C_{4}x^{4}+…]$
• $(1+x)^{n}-(1-x)^{n}=2[C_{1}x+C_{3}x^{3}+C_{5}x^{5}+…]$
• The number of terms in the expansion of $(x+a)^{n}+(x-a)^{n}\;is \;\frac{n+2}{2}\;if\; n \;is\;even\;\frac{n+1}{2}\; if \;n \;is \;odd.$
• The number of terms in the expansion of $(x+a)^{n}-(x-a)^{n}\;is \;\frac{n}{2}\;if\; n \;is\;even\;\frac{n+1}{2}\; if \;n \;is \;odd.$

### Properties of Binomial Coefficients

• $C_{0}+C_{1}+C_{2}+…+C_{n}=2^{n}$
• $C_{0}+C_{2}+C_{4}+…=C_{1}+C_{3}+C_{5}+…=2^{n-1}$
• $C_{0}-C_{1}+C_{2}-C_{3}+…+(-1)^{n}.nC_{n}=0$
• $nC_{1}+2.nC_{2}+3.nC_{3}+…+n.nC_{n}=n.2^{n-1}$
• $C_{1}-2C_{2}+3C_{3}-4C_{4}+…+(-1)^{n-1}C_{n}=0$ for $n>1$
• $C_{0}^{2}+C_{1}^{2}+C_{2}^{2}+…Cn^{2}=\frac{(2n)!}{(n!)^{2}}$

Illustration: If $(1+x)^{15}=a_{0}+a_{1}x+…+a_{15}x^{15}$ then

Find the value of $\sum_{r=1}^{15}r.\frac{ar}{a_{r=1}}$

Sol:

$\sum_{r=1}^{15}r.\frac{ar}{a_{r=1}}=1.\frac{a_{1}}{a_{0}}+2.\frac{a_{2}}{a_{1}}+3.\frac{a_{3}}{a_{2}}+…+15.\frac{a_{15}}{a_{14}}$

$=\frac{C_{1}}{C_{0}}+2.\frac{C_{2}}{C_{1}}+3.\frac{C_{3}}{C_{2}}+…+15.\frac{C_{15}}{C_{14}}$

$=15+14+13+…+1=\frac{15(15+1)}{2}=120$

## Terms in the binomial expansion

### General Term in binomial expansion:

We have $(x+y)^{n}=nC_{0}x^{n}+nC_{1}x^{n-1}.y+nC_{2}x^{n-2}.y^{2}+…+nC_{n}y^{n}$

General Term $=T_{r+1}=nC_{r}x^{n-r}.y^{r}$

• General Term in $(1+x)^n$ is $nC_rx^r$
• In the binomial expansion of $(x+y)^n$ the rth term from end is$(n-r+2)^{th}$ term from the beginning.

Illustration: Find the number of terms in $(1-2x+x^{2})^{50}$

Sol:

$(1-2x+x^{2})^{50}=[(1+x)^{2}]^{50}=(1+x)^{100}$

The number of terms $=100+1=101$

Illustration: Find the fourth term from the end in the expansion of $(2x-\frac{1}{x^{2}})^{10}$

Sol:

Required term $=T_{10-4+2}=T_{8}=10C_{7}(2x)^{3}\left ( \frac{-1}{x^{2}} \right )^{7}=-960x^{-11}$

### Middle Term(S) in the expansion of (x+y) n.n

• If n is even then $\left ( \frac{n}{2}+1 \right )$ Term is the middle Term.
• If n is odd then $\left ( \frac{n+1}{2} \right )^{th}$ and $\left ( \frac{n+3}{2} \right )^{th}$ terms are the middle terms.

Illustration: Find the middle term of $\left ( 1-3x+3x^{2}-x^{3} \right )^{2n}$

Sol:

$\left ( 1-3x+3x^{2}-x^{3} \right )^{2n}=\left [ \left ( 1-x \right )^{3} \right ]^{2n}=\left ( 1-x \right )^{6n}$

Middle Term $=\left ( \frac{6n}{2}+1 \right )$ term $=6nC_{3n}\left ( -x \right )^{3n}$

### Determining a Particular Term:

• In the expansion of $\left ( ax^p+\frac{b}{x^q} \right )^n$ the coefficient of xm is the coefficient of $T_{r+1}$ where $r=\frac{np-m}{p+q}$
• In the expansion of $\left ( x+a \right )^n,$
$\frac{T_{r+1}}{T_r}=\frac{n-r+1}{r}.\frac{a}{x}$

### Independent Term

The term Independent of in the expansion of $\left ( ax^p+\frac{b}{x^q} \right )^n.$ is

$T_{r+1}=_{ }^{n}\textrm{C}_ra^{n-r}b^r,$ where $r=\frac{np}{p+q}$ (integer)

Illustration: Find the independent term of x in $\left ( x+\frac{1}{x} \right )^6$

Sol:

$r=\frac{6(1)}{1+1}=3$

The independent term is $6C_3=20$

Illustration: Find the independent term in the expansion of $\left ( \frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-\sqrt{x}} \right)^{10}$

Sol:

$\left ( \frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-\sqrt{x}} \right)^{10}=\left [ \left ( x^{1/3}+1 \right )-\frac{\sqrt{x}+1}{\sqrt{x}} \right ]^{10}=\left ( x^{1/3}+1=1=\frac{1}{\sqrt{x}} \right )^{10}=\left ( x^{1/3}-\frac{1}{\sqrt{x}} \right )^{10}$

$r=\frac{10(1/3)}{1/3+1/2}=4$ ∴ $T_5=10C_4=210.$

### Numerically greatest term in the expansion of (1+x)n:

• If $\frac{(n+1)\left | x \right |}{\left | x \right |+1}=P,$ is a positive integer then Pth term and (P+1)th terms are numerically greatest terms in the expansion of (1+x)n
• If $\frac{(n+1)\left | x \right |}{\left | x \right |+1}=P+F,$ where P is a positive integer and 0 < F < 1 then (P+1)th term is numerically greatest term in the expansion of (1+x)n.

Illustration: Find the numerically greatest term in (1-3x)10 when $x=\frac{1}{2}$

Sol:

$\frac{(n+1)\left | \alpha \right |}{\left | \alpha \right |+1}=\frac{11(3/2)}{3/2+1}=\frac{33}{5}=6.6$

Therefore, $T_7$ is the numerically greatest term.

$T_{6+1}=10C_6.(-3x)^6=10C_6.\left ( \frac{3}{2} \right )^6.$

### Ratio of Consecutive Terms/Coefficients:

Coefficients of xr and xr+1 are nCr-1 and nCr respectively $\frac{nC_r}{nC_{r-1}}=\frac{n-r+1}{r}$

Illustration: If the coefficients of three consecutive terms in the expansion of (1+x)n are in the ratio 1:7:42 then find the value of n.

Sol:

Let (r+1)th, (r+2)th and (r+3)th be the three consecutive terms. Then nCr: nCr+1: nCr+2=1:7:42

Now $\frac{nC_r}{nC_{r+1}}=\frac{1}{7}$

$\frac{n{{C}_{r}}}{n{{C}_{r+1}}}=\frac{1}{7}\Rightarrow \frac{r+1}{n-r}=\frac{1}{7}\Rightarrow n-8r=7\to \left( 1 \right).$And $\frac{n{{C}_{r+1}}}{n{{C}_{r+2}}}=\frac{7}{42}\Rightarrow \frac{r+2}{n-r-1}=\frac{1}{6}\Rightarrow n-7r=13\to \left( 2 \right)$

From $\left( 1 \right)\And \left( 2 \right)n=55$

## Applications of Binomial Theorem

### Finding Remainder using Binomial Theorem

Illustration: Find the remainder when 7103 is divided by 25

Sol:

$\frac{{{7}^{103}}}{25}=\frac{7{{\left( 49 \right)}^{51}}}{25}=\frac{7{{\left( 50-1 \right)}^{51}}}{25}$

= $\frac{7\left( 25K-1 \right)}{25}=\frac{175K-25+25-7}{25}$

= $\frac{25\left( 7K-1 \right)+18}{25}$

∴ The remainder = 18.

Illustration: If the fractional part of the number $\frac{{{2}^{403}}}{15}$ is $\frac{K}{15}$ then find K.

Sol:

$\frac{{{2}^{403}}}{15}=\frac{{{2}^{3}}{{\left( {{2}^{4}} \right)}^{100}}}{15}$ $=\frac{8}{15}{{\left( 15+1 \right)}^{100}}=\frac{8}{15}\left( 15\lambda +1 \right)=8\lambda +\frac{8}{15}$

∵ 8λ is an integer, fractional part $=\frac{8}{15}$

K = 8.

### Finding Digits of a Number

Illustration: Find the last two digits of the number (13)10

Sol:

${{\left( 13 \right)}^{10}}={{\left( 169 \right)}^{5}}={{\left( 170-1 \right)}^{5}}$

= $5{{C}_{0}}{{\left( 170 \right)}^{5}}-5{{C}_{1}}{{\left( 170 \right)}^{4}}+5{{C}_{2}}{{\left( 170 \right)}^{3}}-5{{C}_{3}}{{\left( 170 \right)}^{2}}+5{{C}_{4}}\left( 170 \right)-5{{C}_{5}}$

= $5{{C}_{0}}{{\left( 170 \right)}^{5}}-5{{C}_{1}}{{\left( 170 \right)}^{4}}+5{{C}_{2}}{{\left( 170 \right)}^{3}}-5{{C}_{3}}{{\left( 170 \right)}^{2}}+5\left( 170 \right)-1$

A multiple of $100+5\left( 170 \right)-1=100K+849$

∴ The last two digits are 49.

### Relation Between two Numbers

Illustration: Find the larger of ${{99}^{50}}+{{100}^{50}}$ and ${{101}^{50}}$

Sol:

${{101}^{50}}={{\left( 100+1 \right)}^{50}}={{100}^{50}}+{{50.100}^{49}}+{{25.49.100}^{48}}+…$

⇒ ${{99}^{50}}={{\left( 100-1 \right)}^{50}}={{100}^{50}}-{{50.100}^{49}}+{{25.49.100}^{48}}-….$

⇒ ${{101}^{50}}-{{99}^{50}}=2\left[ {{50.100}^{49}}+25\left( 49 \right)\left( 16 \right){{100}^{47}}+… \right]$

= ${{100}^{50}}+{{50.49.16.100}^{47}}+…>{{100}^{50}}$

∴ ${{101}^{50}}-{{99}^{50}}>{{100}^{50}}$

⇒ ${{101}^{50}}>{{100}^{50}}+{{99}^{50}}$

### Divisibility Test

Illustration: Show that ${{11}^{9}}+{{9}^{11}}$ is divisible by 10.

Sol:

${{11}^{9}}+{{9}^{11}}={{\left( 10+1 \right)}^{9}}+{{\left( 10-1 \right)}^{11}}$

= $\left( 9{{C}_{0}}{{.10}^{9}}+9{{C}_{1}}{{.10}^{8}}+…9{{C}_{9}} \right)+\left( 11{{C}_{0}}{{.10}^{11}}-11{{C}_{1}}{{.10}^{10}}+…-11{{C}_{11}} \right)$

= $9{{C}_{0}}{{.10}^{9}}+9{{C}_{1}}{{.10}^{8}}+…+9{{C}_{8}}.10+1+{{10}^{11}}-11{{C}_{1}}{{.10}^{10}}+…+11{{C}_{10}}.10-1$

= $10\left[ 9{{C}_{0}}{{.10}^{8}}+9{{C}_{1}}{{.10}^{7}}+…+9{{C}_{8}}+11{{C}_{0}}{{.10}^{10}}-11{{C}_{1}}{{.10}^{9}}+…+11{{C}_{10}} \right]$

= 10K, which is divisible by 10.

Formulae:

• The number of terms in the expansion of ${{\left( {{x}_{1}}+{{x}_{2}}+…{{x}_{r}} \right)}^{n}}$ is $\left( n+r-1 \right){{C}_{r-1}}$
• Sum of the coefficients of ${{\left( ax+by \right)}^{n}}$ is ${{\left( a+b \right)}^{n}}$

If $f\left( x \right)={{\left( {{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+….+{{a}_{m}}{{x}^{m}} \right)}^{n}}$ then

• (a) Sum of coefficients $=f\left( 1 \right)$
• (b) Sum of coefficients of even powers of x is $\frac{f\left( 1 \right)+f\left( -1 \right)}{2}$
• (c) Sum of coefficients of odd powers of x is $\frac{f\left( 1 \right)-f\left( -1 \right)}{2}$

## Binomial Theorem for any Index

Let n be a rational number and x be a real number such that $\left| x \right|<1.$ Then

${{\left( 1+x \right)}^{n}}=1+nx+\frac{n\left( n-1 \right)}{2!}{{x}^{2}}+\frac{n\left( n-1 \right)\left( n-2 \right)}{3!}{{x}^{3}}+…+$ $+\frac{n\left( n-1 \right)\left( n-2 \right)…\left( n-r+1 \right)}{r!}{{x}^{r}}+…\infty$

Proof:

Let $f\left( x \right)={{\left( 1+x \right)}^{n}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+…+{{a}_{r}}{{x}^{r}}+…\left( 1 \right)$ $f\left( 0 \right)={{\left( 1+0 \right)}^{n}}=1$

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

$n{{\left( 1+x \right)}^{n-1}}$ $={{a}_{1}}+2{{a}_{2}}x+3{{a}_{3}}{{x}^{2}}+4{{a}_{4}}{{x}^{3}}+…+r{{a}_{r}}{{x}^{r-1}}+…\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)$

Put x=0, we get n=a1

Differentiating (2) w.r.t. x on both sides, we get

$n\left( n-1 \right){{\left( 1+x \right)}^{n-2}}$ $=2{{a}_{2}}+6{{a}_{3}}x+12{{a}_{4}}{{x}^{2}}+…+r\left( r-1 \right){{a}_{r}}{{x}^{r-2}}+…\,\,\,\,\,\,\,\,\,\left( 3 \right)$

Put x=0, we get ${{a}_{2}}=\frac{n\left( n-1 \right)}{2!}$

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

$n\left( n-1 \right)\left( n-2 \right){{\left( 1+x \right)}^{n-3}}=6{{a}_{3}}+24{{a}_{4}}x+…+r\left( r-1 \right)\left( r-2 \right){{a}_{r}}{{x}_{r-3}}+…$

Put x=0, we get ${{a}_{3}}=\frac{n\left( n-1 \right)\left( n-2 \right)}{3!}$

Similarly, we get ${{a}_{4}}=\frac{n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)}{r!}$ and so on

${{a}_{r}}=\frac{n\left( n-1 \right)\left( n-2 \right)…\left( n-r+1 \right)}{r!}$

Putting the values of ${{a}_{0,}}{{a}_{1}},{{a}_{2}},{{a}_{3}},…,{{a}_{r}}$ obtained in (1), we get

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

### Binomial theorem for Rational Index

The number of rational terms in the expression of ${{\left( {{a}^{1/l}}+{{b}^{1/k}} \right)}^{n}}$ is $\left[ \frac{n}{LCM\,\,of\,\,\{l,k\}} \right]$ when none of and is a factor of and when at least one of and is a factor of is $\left[ \frac{n}{LCM\,\,of\,\,\{l,k\}} \right]+1$ where [.] is the greatest integer function.

Illustration: Find the number of irrational terms in ${{\left( \sqrt[8]{5}+\sqrt[6]{2} \right)}^{100}}.$

Sol: ${{T}_{r+1}}=100{{C}_{r}}{{\left( \sqrt[8]{5} \right)}^{100-r}}.{{\left( \sqrt[6]{2} \right)}^{r}}=100{{C}_{r}}.5\frac{100-r}{8}{{.2}^{\frac{r}{6}}}.$

∴ r=12,36,60,84

The number of rational terms = 4

Number of irrational terms =101-4=97

### Binomial Theorem for Negative Index

1. If rational number and $-1<x<1$ then

• $(1+x)^{n}=1+nx+\frac{n(n-1)}{2!}+\frac{n(n-1)(n-2)}{3!}x^{3} +….+\frac{n(n-1)(n-2)…..(n-4+1)}{r!}……\infty$
• $(1-x)^{n}=1-nx+\frac{n(n-1)}{2!}x^{2}-….+(-1)’\frac{n(n-1)(n-2)……(n-r+1)}{r!}x^{r}+…..\infty$
• $(1-x)^{n}=1+nx+\frac{n(n+1)}{2!}x^{2}+\frac{n(n+1)(n+2)}{3!}x^{3}+….+\frac{n(n+1)(n+2)…(n+r-1)}{r!}x^{r}+…\infty$
• $(1+x)^{-n}=1-nx+\frac{n(n+1)}{2!}x^{2}….\frac{n(n+1)(n+2)}{3!}x^{3}+…+(-1)^{r}\frac{n(n+1)(n+2)…(n+r-1)}{r!}x^{r}+…\infty$
• $(1-x)^{-1}=1+x+x^{2}+x^{3}+…+x^{r}+…\infty$
• $(1+x)^{-1}=1-x+x^{2}-x^{3}+…(-1)^{r}x^{r}+…\infty$
• $(1-x)^{-2}=1+2x+3x^{2}-4x^{3}+…+(r+1)x^{r}+…\infty$
• $(1+x)^{-2}=1-2x+3x^{2}-4x^{3}+…+(-1)^{r}(r+1)x^{r}+…\infty$
• $(1-x)^{-3}=1+3x+6x^{2}+10x^{3}+…+\frac{(r+1)(r+2)}{r!}+…\infty$
• $(1+x)^{-3}=1+3x+6x^{2}+10x^{3}+…+(-1)^{r}\frac{(r+1)(r+2)}{r!}+…\infty$

2. Number of terms in $(1+x)^{n}$ is

• $’n+1$ when positive integer.
• Infinite when is not a positive integer &$\left | x \right |<1$

3. First negative term in ${{\left( 1+x \right)}^{{}^{p}/{}_{q}}}$ when $0<x<1,p,q$ are positive integers & ‘p’ is not a multiple of ‘q’ is ${{T}_{\left[ \frac{p}{q} \right]+3}}$

## Multinomial Theorem

Using binomial theorem, we have

${{\left( x+a \right)}^{n}}$

=$\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}\,\,{{x}^{n-r}}\,{{a}^{r}},n\in N}$

= $\sum\limits_{r=0}^{n}{\frac{n!}{\left( n-r \right)!r!}{{x}^{n-r}}{{a}^{r}}}$

= $\sum\limits_{r+s=n}^{n}{\frac{n!}{r!s!}{{x}^{s}}{{a}^{r}},}$ where s = n – r.

This result can be generalized in the following form:

${{\left( {{x}_{1}}+{{x}_{2}}+…+{{x}_{k}} \right)}^{n}}$

= $\sum\limits_{{{r}_{1}}+{{r}_{2}}+…+{{r}_{k}}=n}{\frac{n!}{{{r}_{1}}!{{r}_{2}}!…{{r}_{k}}!}x_{1}^{{{r}_{1}}}x_{2}^{{{r}_{2}}}}…x_{k}^{{{r}_{k}}}$

The general term in the above expansion is

$\frac{n!}{{{r}_{1}}!{{r}_{2}}!{{r}_{3}}!…{{r}_{k}}!}x_{1}^{{{r}_{1}}}x_{2}^{{{r}_{2}}}x_{3}^{{{r}_{3}}}…x_{k}^{{{r}_{k}}}$

The number of terms in the above expansion is equal to the number of non-negative integral solution of the equation.

${{r}_{1}}+{{r}_{2}}+…+{{r}_{k}}=n,$ because each solution of this equation gives a term in the above expansion. The number of such solutions is ${}^{n+k-1}{{C}_{k-1}}.$

PARTICULAR CASES

Case-1: ${{\left( x+y+z \right)}^{n}}=\sum\limits_{r+s+t=n}{\frac{n!}{r!s!t!}{{x}^{r}}{{y}^{s}}{{z}^{t}}}$

The above expansion has ${}^{n+3-1}{{C}_{3-1}}={}^{n+2}{{C}_{2}}$ terms.

Case-2: ${{\left( x+y+z+u \right)}^{n}}=\sum\limits_{p+q+r+s=n}{\frac{n!}{p!q!r!s!}{{x}^{p}}{{y}^{q}}{{z}^{r}}{{u}^{s}}.}$

There are ${}^{n+4-1}{{C}_{4-1}}={}^{n+3}{{C}_{3}}$ terms in the above expansion.

REMARK: The greatest coefficient in the expansion of ${{\left( {{x}_{1}}+{{x}_{2}}+…+{{x}_{m}} \right)}^{n}}$ is $\frac{n!}{{{\left( q! \right)}^{m-r}}{{\left[ \left( q+1 \right)! \right]}^{r}}},$ where q and r are the quotient and remainder respectively when n is divided by m.

Multinomial Expansions

Consider the expansion of ${{\left( x+y+z \right)}^{10}}.$ In the expansion, each term has different powers of x,y, and z and the sum of these powers is always 10.

One of the terms is $\lambda {{x}^{2}}{{y}^{3}}{{z}^{5}}.$ Now, the coefficient of this term is equal to the number of ways $2x’s,3y’s,$ and $5z’s$ are arranged, i.e., $10!\left( 2!3!5! \right).$ Thus,

$(x+y+z)^{10}=\sum \frac{10!}{P_1!P_2!P_3!} x^{P_1}y^{P_2}z^{P_3}$

Where $P_1+P_2+P_3 = 10$ and 0 ≤ $P_1,P_2,P_3$ ≥ 10 In general,

$(x_1+x_2+…x_r)^{n}=\sum \frac{n!}{P_1!P_2!…P_r!} x^{P_1}x^{P_2}…x^{P_r}$

Where $P_1+P_2+P_3+…+P_r$ = n and 0 ≤ $P_1,P_2,…P_r$ ≥ n

Number of Terms in the Expansion of $(x_1+x_2+…+x_r)^{n}$

From the general term of the above expansion, we can conclude that the number of terms is equal to the number of ways different powers can be distributed to $x_1,x_2,x_3….,x_n$ such that the sum of powers is always

Number of non-negative integral solutions of $x_1+x_2+…+x_r = n \; is ^{n+r-1}C_{r-1}$ For example, number of terms in the expansion of $(x+y+z)^{3} \; is\; ^{3+3-1}C_{3-1}=^{5}C_{2} = 10$

As in the expansion, we have terms such as

$x^{0}y^{0}z^{3},x^{0}y^{1}z^{2},x^{0}y^{2}z^{1},x^{0}y^{3}z^{0},x^{1}y^{0}z^{2},x^{1}y^{1}z^{1},x^{1}y^{2}z^{0},x^{2}y^{0}z^{1},x^{2}y^{1}z^{0},x^{3}y^{0}z^{0}.$

Number of terms in $(x+y+z)^{n}\;is\;^{n+3-1}C_{3-1}=^{n+2}C_{2}$.

Number of terms in $(x+y+z+w)^{n}\;is\;^{n+4-1}C_{4-1}=^{n+3}C_{3}$ and so on.

## Problems on Binomial Theorem

Q.1: If the third term in the binomial expansion of ${{\left( 1+{{x}^{\log \,_{2}^{x}}} \right)}^{5}}$ equals 2560, find x.

Sol:

${{T}_{3}}=5{{C}_{2}}.{{\left( {{x}^{\log _{2}^{x}}} \right)}^{2}}=2560$ $\Rightarrow 10.\,{{x}^{2\log _{2}^{x}}}=2560$ $\Rightarrow {{x}^{2\log _{2}^{x}}}=256$ $\Rightarrow {{\left( \log _{2}^{x} \right)}^{2}}=4\Rightarrow \log _{2}^{x}=2\,\,or\,\,-2\Rightarrow x=4\,\,or\,\,\frac{1}{4}$

Q.2: Find the positive value of λ for which the co-efficient of x2 in the expression ${{x}^{2}}{{\left( \sqrt{x}+\frac{\lambda }{{{x}^{2}}} \right)}^{10}}$ is 720.

Sol:

${{x}^{2}}\left[ 10{{C}_{r}}.{{\left( \sqrt{x} \right)}^{10-r}}.{{\left( \frac{\lambda }{{{x}^{2}}} \right)}^{r}} \right]={{x}^{2}}\left[ 10{{C}_{r}}.\,x\frac{10-r}{2}.\,{{\lambda }^{r}}.\,{{x}^{-2r}} \right]$

= ${{x}^{2}}\left[ 10{{C}_{r}}.{{\lambda }^{r}}.x\frac{10-5r}{2} \right]$

Therefore, r = 2

Hence $10{{C}_{2}}.{{\lambda }^{2}}=720\Rightarrow {{\lambda }^{2}}=16\Rightarrow \lambda =\pm 4$

Q.3: The sum of the real values of x for which the middle term in the binomial expansion of ${{\left( \frac{{{x}^{3}}}{3}+\frac{3}{x} \right)}^{8}}$ equals 5670 is?

Sol:

${{T}_{5}}=8{{C}_{4}}\frac{{{x}^{12}}}{81}\times \frac{81}{{{x}^{4}}}=5670$

⇒ $70{{x}^{8}}=5670\Rightarrow x=\pm \sqrt{3}$

Q.4: Let ${{\left( x+10 \right)}^{50}}+{{\left( x-10 \right)}^{50}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+…+{{a}_{50}}{{x}^{50}}$ for all$x\in R$, then $\frac{{{a}_{2}}}{{{a}_{0}}}$ is equal to?

Sol:

${{\left( 10+x \right)}^{50}}+{{\left( 10-x \right)}^{50}}$

⇒ ${{a}_{2}}=2.50{{C}_{2}}{{.10}^{48}}$

⇒ ${{a}_{0}}={{2.10}^{50}}$

⇒ $\frac{{{a}_{2}}}{{{a}_{0}}}=\frac{50{{C}_{2}}}{{{10}^{2}}}=12.25$

Q.5: Find the coefficient of x9 in the expansion of $\left( 1+x \right)\left( 1+{{x}^{2}} \right)\left( 1+{{x}^{3}} \right)….\left( 1+{{x}^{100}} \right)$.

Sol:

x9 can be formed in 8 ways.

i.e., ${{x}^{9}},{{x}^{1+8}},{{x}^{2+7}},{{x}^{3+6}},{{x}^{4+5}},{{x}^{1+3+5}},{{x}^{2+3+4}}$ and coefficient in each case is

∴ Coefficient of ${{x}^{9}}=1+1+1+…8$ times = 8

Q.6: The coefficients of three consecutive terms of ${{\left( 1+x \right)}^{n+5}}$ are in the ratio 5:10:14, find n.

Sol:

Let ${{T}_{r-1}},{{T}_{r}},{{T}_{r+1}}$ are three consecutive terms of ${{\left( 1+x \right)}^{n+5}}$

${{T}_{r-1}}=(n+5){{C}_{r-2}}\,.\,{{\left( x \right)}^{r-2}}$

${{T}_{r}}=(n+5){{C}_{r-1}}\,.\,{{x}^{r-1}}$

⇒ ${{T}_{r+1}}=(n+5){{C}_{r}}\,.\,{{x}^{r}}$

Given $(n+5){{C}_{r-2}}:(n+5){{C}_{r-1}}:\left( n+5 \right){{C}_{r}}=5:10:14$

So, $\frac{(n+5){{C}_{r-2}}}{5}=\frac{(n+5){{C}_{r-1}}}{10}=\frac{(n+5){{C}_{r}}}{14}$

Comparing first two results we have $n-3r=-9\to \left( 1 \right)$

Comparing last two results we have $5n-12r=-30\to \left( 2 \right)$

From $\left( 1 \right)\And \left( 2 \right)n=6$.

Q.7: The digit in the units place of the number $183!+{{3}^{183}}$.

Sol:

${{3}^{183}}={{\left( {{3}^{4}} \right)}^{45}}{{.3}^{3}}\Rightarrow$ unit digit = 7

183! Ends with 0

∴ Units digit of $183!+{{3}^{183}}$ is 7.

Q.8: Find the total number of terms in the expansion of ${{\left( x+a \right)}^{100}}+{{\left( x-a \right)}^{100}}$.

Sol:

${{\left( x+a \right)}^{100}}+{{\left( x-a \right)}^{100}}=2\left[ 100{{C}_{0}}.{{x}^{100}}{{C}_{2}}\,{{x}^{98}}.\,{{a}^{2}}+…+100{{C}_{100}}\,{{a}^{100}} \right]$

∴ Total Terms = 51.

Q.9: Find the coefficient of t4 in the expansion of ${{\left( \frac{1-{{t}^{6}}}{1-t} \right)}^{3}}$.

Sol:

${{\left( 1-{{t}^{6}} \right)}^{3}}{{\left( 1-t \right)}^{-3}}$

= $\left( 1-{{t}^{18}}-3{{t}^{6}}+3{{t}^{12}} \right){{\left( 1-t \right)}^{-3}}$

Coefficient of in ${{\left( 1-t \right)}^{-3}}=3+4-1{{C}_{4}}=6{{C}_{2}}=15$

The Coefficient of xr in ${{\left( 1-x \right)}^{-n}}=(r+n-1){{C}_{r}}$.

Q.10: Find the ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of ${{\left( {{2}^{1/3}}+\frac{1}{{{2.3}^{1/3}}} \right)}^{10}}$

Sol:

$\frac{{{T}_{5}}}{T_{5}^{1}}=\frac{10{{C}_{4}}{{\left( {{2}^{1/3}} \right)}^{10-4}}{{\left[ \frac{1}{2{{\left( 3 \right)}^{1/3}}} \right]}^{4}}}{10{{C}_{4}}{{\left( \frac{1}{2\left( {{3}^{1/3}} \right)} \right)}^{10-4}}.{{\left( {{2}^{1/3}} \right)}^{4}}}=4.{{\left( 36 \right)}^{1/3}}$

Q.11: Find the coefficient of ${{a}^{3}}{{b}^{2}}{{c}^{4}}d$ in the expansion of ${{\left( a-b-c+d \right)}^{10}}$.

Sol:

Expand ${{\left( a-b-c+d \right)}^{10}}$ using multinomial theorem and by using coefficient property we can obtain the required result.

Using multinomial theorem, we have

${{\left( a-b-c+d \right)}^{10}}=\sum\limits_{{{r}_{1}}+{{r}_{2}}+{{r}_{3}}+{{r}_{4}}=10}{\frac{\left( 10 \right)!}{{{r}_{1}}!{{r}_{2}}!{{r}_{3}}!{{r}_{4}}!}}{{\left( a \right)}^{{{r}_{1}}}}{{\left( -b \right)}^{r2}}{{\left( -c \right)}^{r3}}{{\left( d \right)}^{{{r}_{4}}}}$

We want to get coefficient of ${{a}^{3}}{{b}^{2}}{{c}^{4}},$ this implies that ${{r}_{1}}=3,{{r}_{2}}=2,{{r}_{3}}=4,{{r}_{4}}=1$

∴ Coefficient of ${{a}^{3}}{{b}^{2}}{{c}^{4}}d$ is $\frac{\left( 10 \right)!}{3!2!4}{{\left( -1 \right)}^{2}}{{\left( -1 \right)}^{4}}=12600$

Q.12: Find the coefficient of in the expansion of$(1+x+x^{2}+x^{3})^{11}$.

Sol:

By expanding given equation using expansion formula we can get the coefficient x4

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

⇒ $(1+x+x^{2}+x^{3})^{11}=(1+x)^{11}(1+x^{2})^{11}$

= $(1+^{11}C_{1}x^{2}+^{11}C_{2}x^{2}+^{11}C_{3}x^{3}+^{11}C_{4}x^{4}…..)$

= $(1+^{11}C_{1}x^{2}+^{11}C_{2}x^{4}+…..)$

To find term in from the product of two brackets on the right-hand-side, consider the following products terms as

$1\times ^{11}C_{2}x^{4}+^{11}C_{2}x^{2}\times^{11}C_{1}x^{2}+^{11}C_{4}x^{4}$

= $[^{11}C_{2}+^{11}C_{2}+^{11}C_{1}+^{11}C_{4}]x^{4}$

⇒ $[55+605+330]x^{4}=990x^{4}$

∴ The coefficient of x4 is 990.

Q.13: Find the number of terms free from radical sign in the expansion of $(\sqrt{5}+\sqrt[4]{n})^{100}$

Sol:

$T_{r+1} = 100C_{r}.5\frac{100-r}{2}.11^{r/4}$

Where r = 0, 1, 2, …., 100

r must be 0, 4, 8, … 100

Number of rational terms = 26

Q.14: Find the degree of the polynomial

$[x+(\sqrt{x^{3-1}})^{1/2})]^{5}+[(x-(\sqrt{x^{3-1}})^{1/2})]^{5}$

Sol:

$(x+\sqrt{x^{3-1}})^{5}+(x-\sqrt{x^{3-1}})^{5}$

= $2[5C_{0}x^{5}+5C_{2}x^{5}(x^{3}-1)+5C_{4}x({x^{3}-1})^{2}]$

Highest power = 7.

Q.15: Find the last three digits of 2726

Sol:

By reducing 2726 into the form (730 – 1)n and using simple binomial expansion we will get required digits.

We have $27^{2}=729$

Now $27^{26} = (729)^{13} = (730-1)^{13}$

= $^{13}C_{0}(730)^{13}\;-\;^{13}C_{1}(730)^{12}+^{13}C_{2}(730)^{11}$ – . . . . . . –

$^{13}C_{10}(730)^{3}\;+\;^{13}C_{12}(730)^{2}\;-\;^{13}C_{12}(730)+1$

= $1000m+\frac{(13)(12)}{2}(14)^{2}-(13)(730)+1$

Where ‘m’ is a positive integer

= 1000m + 15288 – 9490 = 1000m + 5799

Thus, the last three digits of 17256 are 799.

#### Practise This Question

The number of distinct terms in the expansion of (x+y2)13+(x2+y)14 is