Prove the identity sum_k=1^n k4 = (n(n+1)(2n+1)(3n2+3n-1))/(30) using mathematical induction.

May 20, 2026Mathematics

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May 20, 2026

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To prove the identity $\sum_{k=1}^{n} k^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$ by mathematical induction, we follow these steps:

Step 1: Base Case ( $n=1$ ) Let $P(n)$ be the statement $\sum_{k=1}^{n} k^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$ . For $n=1$ : LHS: $\sum_{k=1}^{1} k^4 = 1^4 = 1$ . RHS: $\frac{1(1+1)(2 \cdot 1+1)(3 \cdot 1^2+3 \cdot 1-1)}{30} = \frac{1(2)(3)(3+3-1)}{30}

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To prove the identity _k=1^n k^4 = (n(n+1)(2n+1)(3n^2+3n-1))/(30) by mathematical induction, we follow these steps: Step 1: Base Case (n=1) Let P(n) be the statement _k=1^n k^4 = (n(n+1)(2n+1)(3n^2+3n-1))/(30).

Asked on May 20, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

ScanSolve AI AnswerAnswered on May 20, 2026

To prove the identity _k=1^n k^4 = (n(n+1)(2n+1)(3n^2+3n-1))/(30) by mathematical induction, we follow these steps: Step 1: Base Case (n=1) Let P(n) be the statement _k=1^n k^4 = (n(n+1)(2n+1)(3n^2+3n-1))/(30). For n=1: LHS: _k=1^1 k^4 = 1^4 = 1. RHS: $(1(1+1)(2 · 1+1)(3 · 1^2+3 · 1-1))/(30) = (1(2)(3)(3+3-1))/(30)