Binomial Theorem And Its Simple Applications Question 160
Question: If $ {{(1+x-2x^{2})}^{6}}=1+a_1x+a_2x^{2}+….+a_{12}x^{12} $ , then the expression $ a_2+a_4+a_6+….+a_{12} $ has the value
[RPET 1986, 99; UPSEAT 2003]
Options:
A) 32
B) 63
C) 64
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
- $ {{(1+x-2x^{2})}^{6}}=1+a_1x+a_2x^{2}+….+a_{12}x^{12} $ .
Putting $ x=1 $ and $ x=-1 $ and adding the results gives 64 = 2(1+a^2+a^4+…)
$ \therefore ,a_2+a_4+a_6+….+a_{12}=31 $ .
BETA
