Binomial Theorem And Its Simple Applications Question 190
Question: $ \sum\limits_{k=0}^{10}{^{20}C_{k}=} $
[Orissa JEE 2004]
Options:
A) $ 2^{19}+\frac{1}{2}{{,}^{20}}C_{10} $
B) $ 2^{19} $
C) $ ^{20}C_{10} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- $ \sum\limits_{K=0}^{10}{^{20}C_{k}} $ i.e., $ ^{20}C_0+{{,}^{20}}C_1+……+{{,}^{20}}C_{10} $
We know that, $ {{(1+x)}^{n}}={{,}^{n}}C_0+{{,}^{n}}C_1x^{1}+{{,}^{n}}C_2x^{2}+….+{{,}^{n}}C_{n}.x^{n} $
Put $ x=1 $ ; $ 2^{n}={{,}^{n}}C_0+{{,}^{n}}C_1+{{,}^{n}}C_2+…..+{{,}^{n}}C_{n} $
Put $ n=20 $ ; $ 2^{20}={{,}^{20}}C_0+{{,}^{20}}C_1+{{,}^{20}}C_2+……+{{,}^{20}}C_{20} $
$ 2^{20}+,{{,}^{20}}C_{10}=2,[{{,}^{20}}C_0+{{,}^{20}}C_1+……+{{,}^{20}}C_{10}] $
$ {{[}^{20}}C_0+{{,}^{20}}C_1+……+{{,}^{20}}C_{10}]=2^{19}+\frac{1}{2}{{,}^{20}}C_{10} $
$ \sum\limits_{k=0}^{10}{^{20}C_{k}}=2^{19}+\frac{1}{2}{{,}^{20}}C_{10} $ .
BETA
