Binomial Theorem And Its Simple Applications Question 86
Question: If $ C_{r} $ stands for $ ^{n}C_{r} $ , the sum of the given series $ \frac{2(n/2)!(n/2)!}{n!}[C_0^{2}-2C_1^{2}+3C_2^{2}-…..+{{(-1)}^{n}}(n+1)C_n^{2}] $ , Where n is an even positive integer, is
[IIT 1986]
Options:
A) 0
B) $ {{(-1)}^{n/2}}(n+1) $
C) $ {{(-1)}^{n}}(n+2) $
D) $ {{(-1)}^{n/2}}(n+2) $
Show Answer
Answer:
Correct Answer: D
Solution:
- We have $ C_0^{2}-2C_1^{2}+3C_2^{2}-…..+{{(-1)}^{n}}(n+1)C_n^{2} $
$ =[C_0^{2}-C_1^{2}+C_2^{2}-….+{{(-1)}^{n}}C_n^{2}]- $
$ [C_1^{2}-2C_2^{2}+3C_3^{2}….-{{(-1)}^{n}}nC_n^{2}] $ = $ {{(-1)}^{n/2}}\frac{n!}{(n/2)!(n/2)!}-{{(-1)}^{(n/2)-1}}.\frac{1}{2}n,{{,}^{n}}{C_{n/2}} $ = $ {{(-1)}^{n/2}}.\frac{n!}{(n/2)!(n/2)!}.( 1+\frac{n}{2} ) $
Therefore the value of the given expression is $ \frac{2(n/2),!,(n/2)!}{n!}\times {{(-1)}^{n/2}}.\frac{(n)!}{(n/2)!(n/2)!}( 1+\frac{n}{2} ) $
$ ={{(-1)}^{n/2}}(2+n) $
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