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Q.

State and prove binomial theorem 
 

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Detailed Solution

Let n be a positive integer and x, a be real numbers, then 

(x+a)n=nC0.xna0+nC1.xn1a1+nCn.xn2a2+.........+nCr.xnrar+..... +nCn.x0an

We prove this theorem by using the principle of mathematical induction (on n). 
When 
n=1,(x+a)n=(x+a)1=x+a=1C0x1a0+1C1x0a1

Thus the theorem is true for n = 1
Assume that the theorem is true for n = k1
(where k is a positive integer). That is 

(x+a)k=kC0.xka0+kC1.xk1a1+kC2.xk2a2+.........+kCr.xkrar+..... +kCk.x0ak

Now we prove that the theorem is true when n = k + 1 also 

(x+a)k+1=(x+a)(x+a)k

=(x+a)[kC0.xka0+kC1.xk1a1+kC2.xk2a2+.........+kCr.xkrar+.....+kC k.x0ak]=kC0xk+1+kC1xka1+kC2xk1a2+.........+kCrxkr+1ar+.....+kC k.x1ak

=kC0xk.a1+kC1xk1a2+kC2xk2a3+.........+kCrxkrar+1+.....+kC kak+1

=kC0.xk+1.a0+(kC1+kC0).xk.a1+(kC2+kC1).xk1.a2+........+(kCr+kCr1).xkr+1.ar+......+(kCk+kCk1).x1.ak+ kCk.x0.ak+1

Since  kC0=1=k+1C0,kCr+kCr-1=k+1Cr

for 1rk kCk=1=k+1Ck+1

x+ak+1=k+1C0.xk+1.a0+k+1C1.xk.a1+k+1C2.xk-1.a2+ ......k+1Cr.xk-r+1.ar+...+k+1Ck.x1.ak+k+1Ck+1.x0.ak+1

Therefore the theorem is true for n = k +1 Hence, by mathematical induction, it follows
that the theorem is true of all positive integer n: 
 

 




 


 

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