Question

Prove that$\sum _{j=1}^n j^4=n(n+1)(2n+1)\left(3n^2+3n-1\right)/30$ whenever n is a positive integer.

Short answer

$\sum _{j=1}^n j^4=n(n+1)(2n+1)\left(3n^2+3n-1\right)/30$

Step-by-step solution

Step: 1

Let P(n) be the statement that $1^4+2^4+3^4+\dots +n^4=n(n+1)(2n+1)\left(3n^2+3n-1\right)/30$

P (1) is true because 1.2.3.5/30=1.

Step: 2

Assume that P (k) is true.

Then

$\begin{array}{r}{1}^4+2^4+3^4+\dots +k^4+(k+1{)}^4\\ =k(k+1)(2k+1)\left(3k^2+3k-1\right)/30+(k+1{)}^4\\ =\left[\right(k+1)/30]\left[k(2k+1)\left(3k^2+3k-1\right)+30(k+1{)}^3\right]\\ =\left[\right(k+1)/30]\left(6k^4+39k^3+91k^2+89k+30\right)\\ =\left[\right(k+1)/30](k+2)(2k+3)\left[3(k+1{)}^2+3(k+1)-1\right]\end{array}$

This demonstrates that P(k+1) is true.