Algebra
Binomial Theorem
MMTS_Full_Test_22
Grade 12
Question:
For $x\in\mathbb{R}$, $x\ne -1$, $(1+x)^{2016}+x(1+x)^{2015}+x^2(1+x)^{2014}+\cdots+x^{2016}=\sum_{i=0}^{2016}a_i x^i$. Then $a_{17}=$
$\binom{2017}{17}$
$\binom{2016}{16}$
$\binom{2016}{17}$
$\binom{2017}{17}$
Step-by-Step Solution
Key Concept: Sum of GP: $(1+x)^{2017}-x^{2017})/1=\sum a_i x^i$
The series sums to $\frac{(1+x)^{2017}-x^{2017}}{1}$. Coefficient of $x^{17}$ is $\binom{2017}{17}$.
Correct Answer: 1