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Expert-verified Found in: Page 330 ### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095 # Prove that $\sum _{j=1}^{n} {j}^{4}=n\left(n+1\right)\left(2n+1\right)\left(3{n}^{2}+3n-1\right)/30$ whenever n is a positive integer.

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

See the step by step solution

## Step: 1

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

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

## Step: 2

Assume that P (k) is true.

Then

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

This demonstrates that P(k+1) is true. ### Want to see more solutions like these? 