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Q17E

Expert-verifiedFound in: Page 330

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

Prove that $\sum _{j=1}^{n}\u200a{j}^{4}=n(n+1)(2n+1)\left(3{n}^{2}+3n-1\right)/30$ whenever n is a positive integer.

$\sum _{j=1}^{n}\u200a{j}^{4}=n(n+1)(2n+1)\left(3{n}^{2}+3n-1\right)/30$

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

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

Assume that P (k) is true.

Then

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

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

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