Binomial Theorem

Binomial Theorem

• An algebraic expression containing two terms is called binomial expression.
  Example
  $( x+a)$
  $(\frac {1}{2} +x)$
  $(\frac {2}{x} - \frac{1}{x^3})$
  
• The general  form of the binomial expression is  $(x+a)$ and expansion $(x+a)^n , n \in N$ is called Binomial expression
• It was developed by Sir Issac newton
• The general expression for the Binomial Theorem is
  $(x+a)^n = ^{n}C_{0} x^n a^0 + ^{n}C_{1} x^{n-1} a^1 + ^{n}C_{2} x^{n-2} a^2 + .....+ ^{n}C_{r} x^{n-r} a^r ....+^{n}C_{n} x^{0} a^n $
  $(x+a)^{n} = \sum_{k=0}^{n} {}^{n} C_k x^{n-k}a^{k} $
  Proof:
  We can prove this theorem with the help of mathematical induction
  Let us assume P(n) be the statement is
  $(x+a)^n = ^{n}C_{0} x^n a^0 + ^{n}C_{1} x^{n-1} a^1 + ^{n}C_{2} x^{n-2} a^2 + .....+ ^{n}C_{r} x^{n-r} a^r ....+ ^{n}C_{n} x^{0} a^n $
  Step 1
  Now the value of P(1)
  $(x+a)^1 = ^{1}C_{0} x^1 a^0 + ^{1}C_{1} x^{1-1} a^1$
  $=(x+a)$
  So P(1) is true
  Step 2
  Now the value of P(m)
  $(x+a)^m = ^{m}C_{0} x^m a^0 + ^{m}C_{1} x^{m-1} a^1 + ^{m}C_{2} x^{m-2} a^2 + ...+ ^{m}C_{m} x^{0} a^m $
  Now we have to prove
  $(x+a)^{m+1} = ^{m+1}C_{0} x^{m+1} a^0 + ^{m+1}C_{1} x^{m} a^1 + ^{m+1}C_{2} x^{m-1} a^2 +...+ ^{m+1}C_{m+1} x^{0} a^{m+1} $
  Now
  $(x+a)^{m+1} =(x+a)(x+a)^m $
  $= (x+a)( ^{m+1}C_{0} x^{m+1} a^0 + ^{m+1}C_{1} x^{m} a^1 + ^{m+1}C_{2} x^{m-1} a^2 + ...+ ^{m+1}C_{m+1} x^{0} a^{m+1} $
  $= ^m C_0 x^{m+1}a{0} + ( ^mC_1 + ^mC_0) x^ma^1 + (^mC_2 + ^mC_1) x^{m-1}a^2 +.... $ $+(^m C_{m-1} + ^mC_{m}) x^1a^m + ^mC_m x^0a^{m+1}$
  As $ ^mC_{r-1} + ^mC_r = ^{m+1}C_r$
  So,
  $= ^{m+1}C_{0} x^{m+1} a^0 + ^{m+1}C_{1} x^{m} a^1 + ^{m+1}C_{2} x^{m-1} a^2 + ...+ ^{m+1}C_{m+1} x^{0} a^{m+1} $
  So by principle of Mathematical induction, P(n) is true for $n \in N$
  

Important conclusion from Binomial Theorem

1	$(x+a)^n =\sum_{k=0}^{n} {}^{n} C_k x^{n-k}a^{k} $ We can easily see that $(x+a)^n$ has $(n+1)$ terms as k can have values from 0 to n
2	The sum of indices of x and a in each is equal to n $x^{n-k}a^{k}$
3	The coefficient nCr is each term is called binomial coefficient
4	$(x-a)^n$ can be treated as $[x+(-a)]^n$ So $(x -a)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{n-k}a^k $ So the terms in the expansion are alternatively positive and negative. The  last term is positive or negative depending on the values of n
5.	$(1+x)^n$ can be treated as $(x+a)^1$ where x=1 and a=x So $(1 +x)^n =\sum_{k=0}^{n} {}^{n} C_k x^{k}$ This is the expansion is ascending order of  power of x
6.	$(1+x)^n$ can be treated as $(x+a)^n$ where x=x and a=1 So $(1 +x)^n =\sum_{k=0}^{n} {}^{n} C_k x^{n-k}$ This is the expansion is descending  order of  power of x
7.	$(1-x)^n$ ;can be treated as $(x+a)^n$ where x=1 and a=-x So $(1 -x)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{k}$ This is the expansion is ascending order of  power of x

General Term in Binomial Expansion

$(x+a)^n =\sum_{k=0}^{n} {}^{n} C_k x^{n-k}a^k $
The First term would be = $ ^{n} C_0 x^n a^0$
The Second term would be =$ ^{n} C_1 x^{n-1} a^1$
The Third term would be = $ ^{n} C_2 x^{n-2} a^2$
The Fourth term would be = $ ^{n} C_3 x^{n-3} a^3$
Like wise (k+1) term would be
$T_{k+1}= ^n C_k x^{n-k}a^k $
This is called the general term also as every term can be find using this term
$T_1= T_{0+1}=^n C_0 x^n a^0 $
$T_2= T_{1+1}=^n C_1 x^{n-1} a^1 $
Similarly for
$(x -a)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{n-k}a^k $
$T_{k+1}= ^n C_k (-1)^k x^{n-k}a^k $
Again similarly for
$(1 +x)^n =\sum_{k=0}^{n} {}^{n} C_k x^{k}$
$T_{k+1}= ^n C_k x^{k}$
Again similarly for
$(1 -x)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{k}$
$T_{k+1}= ^n C_k (-1)^k x^{k}$
To summarize it

Binomial term	(k+1) term
$(x+a)^n =\sum_{k=0}^{n} {}^{n} C_k x^{n-k}a^k $	$T_{k+1}= ^n C_k x^{n-k}a^k $
$(x -a)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{n-k}a^k $	$T_{k+1}= ^n C_k (-1)^k x^{n-k}a^k $
$(1 +x)^n =\sum_{k=0}^{n} {}^{n} C_k x^{k}$	$T_{k+1}= ^n C_k x^{k}$
$(1 -x)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{k}$	$T_{k+1}= ^n C_k (-1)^k x^{k}$

Middle Term in Binomial Expansion

• A binomial expansion contains $(n+1)$ terms
• If n is even then the middle term would $[(\frac {n}{2}+1 ]$ th term
• If n is odd,then $\frac{n+1}{2} $ and $ \frac {n+3}{3}$ are the middle term

Related Topics

Class 12 Maths Class 12 Physics