Short Answer

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

$\sum_{j = 1}^{n} j^{4} = n (n + 1) (2 n + 1) (3 n^{2} + 3 n - 1) / 30$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step: 1

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

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

Step: 2

Assume that P (k) is true.

Then

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

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

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Discrete Mathematics and its Applications

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Short Answer

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

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