Talk:Sum of Sequences of Fifth and Seventh Powers

Why don't you use induction to first prove

sum n5 = k2 (k+1)2 (2k2+2k-1)/12 = 2 k2 (k+1)2 (2k2+2k-1) / 24

sum n7 = k2 (k+1)2 (3k4+6k3-k2-4k+2)/24

then it follows that

sum n5 + n7 = k2(k+1)2 (4k2+4k-2 + 3k4+6k3-k2-4k+2) /24

= k2(k+1)2 (3k2+3k4+6k3) /24

= 3 k4 (k+1)4 / 24

= 2 (k (k+1)/ 2)4

= 2 (sum n)4

(Maybe you'll find what you need here http://mathworld.wolfram.com/PowerSum.html)

MattH (talk)


 * Please feel free to edit the wiki according to your recommendations. You are alerted to the house style, and you are invited to use $\LaTeX$ (the MathJax dialect). Tutorials are available online, but in this context it is straightforward (and indeed preferable) to learn by following the examples of already-posted content. --prime mover (talk) 05:43, 1 May 2017 (EDT)