Binomial Theorem And Its Simple Applications Question 34
Question: The sum of the series $ ^{20}C_0-{{,}^{20}}C_1+{{,}^{20}}C_2-{{,}^{20}}C_3+…. $ $ -….+{{,}^{20}}C_{10} $ is
Options:
A) 0
B) $ ^{20}C_{10} $
C) $ {{-}^{20}}C_{10} $
D) $ \frac{1}{2}{{,}^{20}}C_{10} $
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Answer:
Correct Answer: D
Solution:
- [d] We know that, $ {{(1+x)}^{20}}={{,}^{20}}C_0+{{,}^{20}}C_1x+ $
$ ^{20}C_2x^{2}+….{{,}^{20}}C_{10}x^{10}+….{{,}^{20}}C_{20}x^{20} $
Put $ x=-1,,(0)={{,}^{20}}C_0-{{,}^{20}}C_1+{{,}^{20}}C_2-{{,}^{20}}C_3+…. $
$ {{+}^{20}}C_{10}-{{,}^{20}}C_{11}…+{{,}^{20}}C_{20} $
$ \Rightarrow 0=2{{[}^{20}}C_0-{{,}^{20}}C_1+{{,}^{20}}C_2-{{,}^{20}}C_3+….-{{,}^{20}}C_9] $
$ +{{,}^{20}}C_{10} $
$ \Rightarrow {{,}^{20}}C_{10}=2{{[}^{20}}C_0-{{,}^{20}}C_1+{{,}^{20}}C_2-{{,}^{20}}C_3 $
$ +….-{{,}^{20}}C_9+{{,}^{20}}C_{10}] $
$ \Rightarrow {{,}^{20}}C_0-{{,}^{20}}C_1+{{,}^{20}}C_2-{{,}^{20}}C_3+…+{{,}^{20}}C_{10} $
$ =\frac{1}{2}{{,}^{20}}C_{10} $
BETA
