Binomial Theorem And Its Simple Applications Question 176
Question: If $ {{(1+x)}^{n}}=C_0+C_1x+C_2x^{2}+…+C_{n}x^{n} $ , then the value of $ C_0+C_2+C_4+C_6+….. $ is
[RPET 1997]
Options:
A) $ {2^{n-1}} $
B) $ {2^{n-1}} $
C) $ 2^{n} $
D) $ {2^{n-1}}-1 $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ {{(1+x)}^{n}}=C_0+C_1x+C_2x^{2}+….+C_{n}x^{n} $ Putting x = 1, we get
$ \Rightarrow 2^{n}=C_0+C_1+C_2+…..+C_{n} $ …..(i) or $ C_1+C_2+C_3+….+C_{n}=2^{n}-1 $
$ [\because ,C_0={{,}^{n}}C_0=1] $
Again, putting x = -1, we get $ 0=C_0-C_1+C_2-C_3+…. $ or $ C_0+C_2+C_4+…. $
$ =C_1+C_3+C_5+…. $ i.e. $ A=B $ Also from (i), A + B = 2n or $ A={2^{n-1}}=B $
Hence, $ C_0+C_2+C_4+….={2^{n-1}} $ .
BETA
