Binomial Theorem
Sum of alternate coefficients via roots of unity
MJMT_Full_Test_06
Grade 12
Question:
If $(1+x)^{2010}=C_0+C_1x+\cdots+C_{2010}x^{2010}$, then $C_2+C_5+C_8+\cdots+C_{2009}$ equals
$\dfrac{1}{2}(2^{2010}-1)$
$\dfrac{1}{3}(2^{2010}-1)$
$\dfrac{1}{2}(2^{2009}-1)$
$\dfrac{1}{3}(2^{2009}-1)$
Step-by-Step Solution
Key Concept: Put $x=1,\omega,\omega^2$ in $(1+x)^{2010}$ and add with appropriate roots of unity factors. $S_0+S_1+S_2=2^{2010}$, $S_0=(2^{2010}+2)/3$, $S_1=S_2=(2^{2010}-1)/3$. The required sum is $S_2=(2^{2010}-1)/3$.
$C_2+C_5+\cdots+C_{2009}=\dfrac{2^{2010}-1}{3}$.
Correct Answer: 2