Find the coefficient of $x^{26}$ in $(x^2+x^3+x^4+x^5+x^6)^8$.

Question

Find the coefficient of $x^{26}$ in $(x^2+x^3+x^4+x^5+x^6)^8$.

my attempt

$(x^2+x^3+x^4+x^5+x^6)^8=x^{16}(1+x+x^2+x^3+x^4)^8=x^{16}((1+x)(1+x^2)+x^4)^8$.

Answer 1

\begin{align}[x^{16}](1-x)^{-8}(1-x^8)^5 &=[x^{16}] \left( 1 + {8 \choose 1}x + {9 \choose 2 }x^2 + \cdots \right)\left(1 - {8 \choose 1}x^5 + {8 \choose 2}x^{10} + \cdots\right)\\&={8 \choose 2} - {8 \choose 1} {12 \choose 5}+ {17 \choose 10}. \end{align}

Note:

$$(1-x^5)^8=\sum_{k=0}(-1)^k\binom{8}{k}x^{5k}$$

$$(1-x)^{-8}=\sum_{k=0}(-1)^k\binom{-8}{k}x^k=\sum_{k=0}\binom{8+k-1}{k}x^k$$

Answer 2

Try to use $$ (x^2+x^3+x^4+x^5+x^6)^8=x^{16}\frac{(1-x^5)^8}{(1-x)^8} $$ and use the generalized geometric/binomial/Newton series $$ (1-x)^{-a}=\sum_{k=0}^\infty\binom{-a}{k}\,(-x)^k=\sum_{k=0}^\infty\binom{k+a-1}{k}\,x^k. $$