# Binomial Theorem

## Binomial Theorem

An algebraic expression containing two terms is called binomial expression.
Example
( x+a)
(1/2 +x)
(2/x -1/x3)
The general  form of the binomial expression is  (x+a)  and expansion (x+a)n  n? N  is called Binomial expression
It was developed by Sir Issac newton
The general expression for the Binomial Theorem is
(x+a)n= nC0 xna0  + nC1 xn-1a1   + nC2 xn-2a2    +………………+ nCr xn-rar …….+  nCn x0an  Proof:
We can prove this theorem with the help of mathematical induction
Let us assume  P(n) be the statement is
(x+a)n= nC0 xna0  + nC1 xn-1a1   + nC2 xn-2a2    +………………+ nCr xn-rar …….+  nCn x0an  Step 1
Now the value of P(1)
(x+a)1 = 1C0 x1a0  + 1C1 x1-1a1
=(x+a)
So P(1) is true
Step 2
Now the value of P(m)
(x+a)m = mC0 xma0  + mC1 xm-1a1   + mC2 xm-2a2    +………………+ mCr xm-rar …….+  mCn x0am          Now we have to prove
(x+a)m+1 = m+1C0 xm+1a0  + m+1C1 xma1   + m+1C2 xm-1a2    +………………+ m+1Cr xm-r-1ar …….+  m+1Cm+1 x0am+1  Now
(x+a)m+1 =(x+a)(x+a)m
= (x+a)(mC0 xma0  + mC1 xm-1a1   + mC2 xm-2a2    +………………+ mCr xm-rar …….+  mCn x0am)
= mC0 xm+1a0  + (  mC1 + mC0) xma1  +   (  mC2 + mC1) xm-1a2  +   ………
+(  mCm-1 + mCm) x1am +   mCm x0am+1
As   mCr-1 + mCr = m+1Cr
So
= m+1C0 xm+1a0  + m+1C1 xma1   + m+1C2 xm-1a2    +………………+ m+1Cr xm-r-1ar …….+  m+1Cm+1 x0am+1  So by principle of Mathematical induction, P(n) is true for n? N

## Important conclusion from Binomial Theorem

 1 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 xn-kak 3 The coefficient nCr is each term is called binomial coefficient 4 (x-a)n  can be treated as [x+(-a)]n So 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)n  where x=1 and a=x So 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 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 This is the expansion is ascending order of  power of x
Examples: Expand using Binomial Theorem

(x2 +2)5 Solution: We know from binomial Theorem
(x+a)n= nC0 xna0  + nC1 xn-1a1   + nC2 xn-2a2    +………………+ nCr xn-rar …….+  nCn x0an  So putting values x=x2  ,a=2  and n=5
We get
(x2 +2)5  = 5C0 (x2)5  + 5C1 (x2)421   + 5C2 (x2)322    +5C3 (x2)223    +5C4 (x2)124    +5C5 (x2)025    =x10 +  20x8 +160x6 +640x4 +1280x2 +1024

## General Term in Binomial Expansion

The First term would be = nC0 xna0
The Second term would be = nC1 xn-1a1
The Third term would be = nC2 xn-2a2
The Fourth term would be = nC3 xn-3a3
Like wise (k+1) term would be
Tk+1= nCk xn-kak
This is called the general term also as every term can be find using this term
T1= T0+1= nC0 xna0
T2= T1+1= nC1 xn-1a1
Similarly for
Tk+1= nCk (-1)k xn-kak
Again similarly for
Tk+1= nCk xk
Again similarly for
Tk+1= nCk  (-1)k  xk
To summarize it
 Binomial term (k+1) term Tk+1= nCk xn-kak Tk+1= nCk (-1)k xn-kak Tk+1= nCk xk Tk+1= nCk  (-1)k  xk

## Middle Term in Binomial Expansion

A binomial expansion contains (n+1) terms
If n is even then the middle term would  [(n/2)+1 ]th term
If n is odd,then (n+1)/2  and (n+3)/3  are the middle term
Examples:
If the coefficient of (2k + 4)  and (k - 2) terms in the expansion of (1+x)24 are equal then find the value of k
Solution:
The general term of (1 + x)n is
Tk+1= nCk xk
Hence coefficient of (2k + 4)th term will be
T2k+4 = T2k+3+1 = 24C2k+3
and coefficient or (k - 2)th term will be
Tk-2 = Tk-3+1 = 24Ck-3
As per question both the terms are equal
24C2k+3 = 24Ck-3.
=> (2k + 3) + (k-3) = 24
.·.     r = 8