Binomial Theorem And Its Simple Applications Question 174
Question: If $ {{(1+x)}^{n}}=C_0+C_1x+C_2x^{2}+….+C_{n}x^{n} $ , then $ C_0C_2+C_1C_3+C_2C_4+{C_{n-2}}C_{n} $ equals
[RPET 1996]
Options:
A) $ \frac{(2n)!}{(n+1)!(n+2)!} $
B) $ \frac{(2n)!}{(n-2)!(n+2)!} $
C) $ \frac{(2n)!}{(n)!(n+2)!} $
D) $ \frac{(2n)!}{(n-1)!(n+2)!} $
Show Answer
Answer:
Correct Answer: B
Solution:
- We have, $ {{(1+x)}^{n}}=C_0+C_1x+C_2x^{2}….+C_{n}x^{n} $
$ {{( 1+\frac{1}{x} )}^{n}}=C_0+C_1.\frac{1}{x}+C_2.\frac{1}{x^{2}}+…+C_{n}( \frac{1}{x^{n}} ) $
on multiplying both expansions, we get $ \frac{{{(1+x)}^{2n}}}{x^{n}}=\sum{C_0^{2}+x\sum{C_0C_1+x^{2}\sum{C_0C_2+….}}} $
$ +x^{r}\sum{C_0C_{r}+…..} $
The various sigma are the coefficient of $ x^{0},x,x^{2},…..,x^{r} $ in L.H.S. $ \frac{{{(1+x)}^{2n}}}{x^{n}} $ or coefficient of $ x^{n},{x^{n+1}},{x^{n+2}},…..,{x^{n+r}} $ in the expansion of $ {{(1+x)}^{2n}} $ which occur in $ {T_{n+1,}}{T_{n+2}},…. $ and are $ ^{2n}C_{n}{{,}^{2n}}{C_{n+1}}{{,}^{2n}}{C_{n+2}}{{….}^{2n}}{C_{n+r}} $ etc. $ ^{,2n}{C_{n+2}}=\frac{(2,n,),!}{(n-2),!,(n+2),!} $
BETA
