Prove that $1^5+2^5+\dots+n^5= \dfrac{n^2(n+1)^2(2n^2+2n-1)}{12}$
I think we use induction to solve this problem.
Base case: When $n = 1$, we have $1^5 = 1$. The formula gives $1$, which is correct.
Inductive hypothesis: Assume that the formula holds for some value of $k,$ where $k$ is a positive integer.
Inductive step: We want to show that the formula also holds for $k+1$.
From here I can't make progress so please help me.
Best Answer
We want to show that $$1^5+2^5+\dots+k^5 +(k+1)^5 = \dfrac{(k+1)^2 (k+2)^2(2(k+1)^2+2(k+1)-1)}{12}.$$ By the inductive hypothesis, we have: $$1^5+2^5+\dots+k^5= \dfrac{k^2(k+1)^2(2k^2+2k+1)}{12}.$$ Now add $(k+1)^5$: \begin{align} 1^5+2^5+\dots+k^5 +(k+1)^5 &= \dfrac{k^2(k+1)^2(2k^2+2k-1)}{12} + (k+1)^5\\ &= \dfrac{k^2(k+1)^2(2k^2+2k-1)}{12} + \dfrac{12(k+1)^5}{12}\\ &= \dfrac{(k+1)^2 (k^2(2k^2+2k-1) + 12(k+1)^3)}{12}\\ &= \dfrac{(k+1)^2 (2k^4+2k^3-k^2 + 12k^3+36k^2+36k+12)}{12}\\ &= \dfrac{(k+1)^2 (\color{red}{2k^4+14k^3+35k^2+36k+12})}{12}. \end{align} Notice that: \begin{align} (k+2)^2(2(k+1)^2+2(k+1)-1)&=(k^2+4k+4)(2k^2+4k+2+2k+1)\\ &=\color{red}{2 k^4 + 14 k^3 + 35 k^2 + 36 k + 12}. \end{align} Therefore $$1^5+2^5+\dots+k^5 +(k+1)^5 = \dfrac{(k+1)^2 (k+2)^2(2(k+1)^2+2(k+1)-1)}{12}.$$ As we wanted to show.