Binomial Theorem And Its Simple Applications Question 67
Question: If $ a_{n}=2n+1 $ and $ C_{r}={{,}^{n}}C_{r} $ then $ a_0C_0^{2}+a_1C_1^{2}+a_2C_2^{2}+……..a_{n}C_n^{2}= $
Options:
A) $ (n-1){{(}^{2n}}C_{n}) $
B) $ n{{(}^{2n}}C_{n}) $
C) $ (n+1){{(}^{2n}}C_{n}) $
D) $ (n+1){{(}^{n}}{C_{n/2}}) $
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] $ S_{n}=a_0C_0^{2}+a_1C_1^{2}+a_2C_2^{2}+…….+a_{n}C_n^{2} $
$ \frac{S_{n}=a_{n}C_n^{2}+{a_{n-1}}{C_{n-1}}^{2}+{a_{n-2}}^{2}+….+a_0C_0^{2}}{2S_{n}=(a_0+a_{n})C_0^{2}+(a_1+{a_{n-1}})C_1^{2}+…+(a_{n}+a_0)C_n^{1}} $
$ =(2n+2)(C_0^{2}+C_1^{2}+C_2^{2}+…..+C_n^{2}) $
$ \therefore S_{n}={{(n+1)}^{2n}}C_{n} $
$ [\because ,a_0+a_{n}=a_1+{a_{n-1}}+…….=2n+2] $
BETA
