Binomial Theorem And Its Simple Applications Question 192
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}= $
Options:
A) $ \frac{(2n)!}{{{(n!)}^{2}}} $
B) $ \frac{(2n)!}{(n-1)!(n+1)!} $
C) $ \frac{(2n)!}{(n-2)!(n+2)!} $
D) none of these
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] $ {{(1+x)}^{n}}=C_0+C_1x-C_2x^{2}+C_3x^{3}…+{C_{n-1}}{x^{n-1}}+C_{n}x^{n}…(1) $
$ {{(1+x)}^{n}}=C_0x^{n}+C_1{x^{n-1}}+C_2{x^{n-2}}+…+{C_{n-1}}x+C_{n}…(2) $
Multiplying Eqs. (1) and (2) and equating the coefficient of $ {x^{\pi -2}} $ , we get $ C_0C_2+C_1C_3+C_2C_4+…+{C_{n-2}}C_{n} $ =Coefficient of $ {x^{n-2}} $ in $ {{({1^{+}}x)}^{2n}} $ = $ ^{2n}{C_{n-2}} $
$ =\frac{(2n)!}{(n-2)!(n+2)!} $
BETA
