Binomial Theorem And Its Simple Applications Question 203
Question: If $ {{(3+x^{2008}+x^{2009})}^{2010}}=a_0+a_1x+a_2x^{2}+….+a_{n}x^{n} $ , then the value of $ a_0-\frac{1}{2}a_1-\frac{1}{2}a_2+a_3-\frac{1}{2}a_4-\frac{1}{2}a_5+a_6-… $ is
Options:
A) $ 3^{2010} $
B) 1
C) $ 2^{2010} $
D) none of these
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] put $ x=\omega ,{{\omega }^{2}} $
$ {{(3+\omega +{{\omega }^{2}})}^{2010}}=a_0+a_1\omega +a_2{{\omega }^{2}}+… $
$ \Rightarrow 2^{2010}=a_0+a_1{{\omega }^{2}}+a_2\omega +a_3+a_2\omega +… $
And $ 2^{2010}=a_0+a_1{{\omega }^{2}}+a_2\omega +a_3+a_2\omega +… $ Adding (1) and (2), we have
$ 2\times 2^{2010}=2a_0-a_1-a_2+2a_3-a_4-a_5+2a_6-… $ Or $ 2^{2010}=a_0-\frac{1}{2}a_1-\frac{1}{2}a_2+a_3-\frac{1}{2}a_4-\frac{1}{2}a_5+{a_{6…}} $
BETA
