Q.

State and prove binomial theorem 
 

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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: 
 

 




 


 

Watch 3-min video & get full concept clarity

hear from our champions

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon