Finding a closed form for $\sum_{k=0}^{n}k^5$ using generating functions

Question

Question:

Find a closed form for $\sum_{k=0}^{n}k^5$ using generating functions.

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$Solution.$

We define $f(x)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}k^5x^n=\sum_{n=0}^{\infty}a_nx^n$.

Therefore, $$\begin{array}{l} a_{n} =\{1,1+32,1+32+243,1+32+243+1024,\dotsc \}\\ \\ b_{n} =\{1,32,243,1024,3125,7776,16807,32768,\dotsc \}\\ \\ c_{n} =\{1,31,211,781,2101,4651,9031,15961,\dotsc \}\\ \\ d_{n} =\{1,30,180,570,1320,2550,4380,6930,\dotsc \}\\ \\ e_{n} =\{1,29,150,390,750,1230,1830,2550,\dotsc \}\\ \\ f_{n} =\{1,28,121,240,360,480,600,720,\dotsc \}\\ \\ g_{n} =\{1,27,93,119,120,120,120,120,\dotsc \}\\ \\ h_{n} =\{1,26,66,26,1,0,0,0,\dotsc \}\\ \\ \Longrightarrow h_{n} =1+26x +66x^{2} +26x^{3} +x^{4}\\ \\ \Longrightarrow a_{n} =\frac{1+26x +66x^{2} +26x^{3} +x^{4}}{( 1-x)^{7}}\\ \\ \Longrightarrow f( x) =\frac{1+26x +66x^{2} +26x^{3} +x^{4}}{( 1-x)^{7}}\\ \\ =\frac{1}{( 1-x)^{7}} +\frac{26x}{( 1-x)^{7}} +\frac{66x^{2}}{( 1-x)^{7}} +\frac{26x^{3}}{( 1-x)^{7}} +\frac{x^{4}}{( 1-x)^{7}}\\ \\ \\ =\sum _{n=0}^{\infty }\binom{n+7-1}{7-1} x^{n} +26\sum _{n=0}^{\infty }\binom{n+7-1}{7-1} x^{n+1} +66\sum _{n=0}^{\infty }\binom{n+7-1}{7-1} x^{n+2}\\ +26\sum _{n=0}^{\infty }\binom{n+7-1}{7-1} x^{n+3} +\sum _{n=0}^{\infty }\binom{n+7-1}{7-1} x^{n+4}\\ \\ =\sum _{n=0}^{\infty }\binom{n+6}{6} x^{n} +26\sum _{n=0}^{\infty }\binom{n+6}{6} x^{n+1} +66\sum _{n=0}^{\infty }\binom{n+6}{6} x^{n+2}\\ +26\sum _{n=0}^{\infty }\binom{n+6}{6} x^{n+3} +\sum _{n=0}^{\infty }\binom{n+6}{6} x^{n+4} \end{array}$$

By substitution, $$ \begin{array}{l} =\sum _{n=0}^{\infty }\binom{n+6}{6} x^{n} +26\sum _{n=0}^{\infty }\binom{n-1+6}{6} x^{n} +66\sum _{n=0}^{\infty }\binom{n-2+6}{6} x^{n}\\ +26\sum _{n=0}^{\infty }\binom{n-3+6}{6} x^{n} +\sum _{n=0}^{\infty }\binom{n-4+6}{6} x^{n}\\ \\ =\sum _{n=0}^{\infty }\binom{n+6}{6} x^{n} +26\sum _{n=0}^{\infty }\binom{n+5}{6} x^{n} +66\sum _{n=0}^{\infty }\binom{n+4}{6} x^{n}\\ +26\sum _{n=0}^{\infty }\binom{n+3}{6} x^{n} +\sum _{n=0}^{\infty }\binom{n+2}{6} x^{n}\\ \\ =\sum _{n=0}^{\infty }\left[\binom{n+6}{6} +26\binom{n+5}{6} +66\binom{n+4}{6} +26\binom{n+3}{6} +\binom{n+2}{6}\right] x^{n}\\ \\ \Longrightarrow a_{n} =\binom{n+6}{6} +26\binom{n+5}{6} +66\binom{n+4}{6} +26\binom{n+3}{6} +\binom{n+2}{6} \end{array}$$

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Now, I have checked over the Wolfram site, and it is incorrect. I don't know where I was wrong.

Answer 1

It seems there's just an off-by-one error. Note that since \begin{align*} \color{blue}{f(x)}&=\sum_{n=0}^\infty\sum_{k=0}^n k^5 x^n=0+x+32x^2+276 x^3+1\,300 x^4\cdots\\ &=\color{blue}{\frac{x+26x^2+66x^3+26x^4+x^5}{(1-x)^7}} \end{align*} the coefficients $a_n$ should be considered as \begin{align*} \begin{array}{c|rrrrrr} n&0&1&2&3&4&\cdots\\ a_n&0&1&33&276&1\,300&\cdots\tag{1} \end{array} \end{align*}

Taking $a_n$ as in (1) will result in \begin{align*} \color{blue}{\sum_{k=1}^nk^5}&=\sum_{m=0}^4A(5,m)\binom{n+m+1}{6}\\ &\,\,\color{blue}{=\binom{n+1}{6}+26\binom{n+2}{6}+66\binom{n+3}{6}}\\ &\,\,\qquad\color{blue}{+26\binom{n+4}{6}+\binom{n+5}{6}} \end{align*} where $A(5,m): 1,26,66,26,1$ are Eulerian numbers which provide nice connections with $k$-th powers of natural numbers.

All other calculations of OP are correct. The sequence \begin{align*} &\binom{n+6}{6} +26\binom{n+5}{6} +66\binom{n+4}{6} \\ &\qquad+26\binom{n+3}{6} +\binom{n+2}{6}\qquad\qquad (n\geq 0) \end{align*} starts with $1,33,276,1\,300, \ldots$ corresponding to OPs start of the calculation.