What is the coefficient of $x^{17}$ in the formula $(x^2+x)^{15} $?

Question

What is the coefficient of $x^{17}$ in the formula $(x^2+x)^{15} $?

Any idea how to solve this using the binomial coefficient formula?

Answer 1

$(x+x^2)^{15} = x^{15} (1+x)^{15}$.

Now look for the coefficient of $x^2$ in $(1+x)^{15}$.

Or, since $(x+x^2)^{15} = \sum_{k=0}^{15} \binom{15}{k} x^{k}x^{2(15-k)}$, look for the coefficient of $17$ which will correspond to $k=13$.

Or, differentiate 17 times and set $x=0$. Then divide by $17!$.

Answer 2

Rewrite it as $x^{15} (1+x)^{15}$, then expand this using Binomial theorem. Can you handle from here?

Answer 3

Hint : ask yourself how summands can contain $x^{17}$.

Answer 4

$$(x^2+x)^{15}=x^{15}(x+1)^{15}=x^{15}\sum_{k=0}^{15}\binom{15}{k}x^k=\sum_{k=0}^{15}\binom{15}{k}x^{k+15}$$