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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                        

 


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