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.

Table of Contents

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.

C:\Users\Krishna\Desktop\PASCAL TRIANGLE.jpg

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