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The objective is to prove the following equation for $n\ge1$:

$1^4 +2^4 +3^4 +...+n^4 = n(n+1)(2n+1)(3n^2 +3n−1)/30$

I have worked for an hour in it and I get a different answer every time, could use a helping hand.

Answer:

The problem. Prove that $1^4 + 2^4 + 3^4 + ... + n^4 = n(n + 1)(2n + 1)(3n^2 + 3n - 1)/30$ holds for all $n\ge1$ using induction.

Base Case. We verify the equality for $n=1$. We have that $1^4 = 1$ and $(1\cdot2\cdot3\cdot5) / 30=1$, so equality holds.

Induction. Assume that the formula is true for $n=k$, so we have that

$$1^4 + 2^4 + 3^4 + \ldots + k^4 =k(k + 1)(2k + 1)(3k^2 + 3k - 1)/30.$$

The goal is to prove now that, under this assumption, the expression

$$k(k + 1)(2k + 1) (3k^2 + 3k - 1)/30 + (k+1)^4 = (k+1)(k+2)(2k + 3)(3(k+1)^2+3k+2)/30$$ holds.

Dividing by $(k+1)$ and multiplying by $30$ on both sides yield

$$k(2k+1)(3k^2 + 3k-1) + 30(k+1)^3 = (k+2)(2k + 3)(3(k+1)^2+3k+2),$$

which after expansion yields

$$6k^4 + 39k^3 + 91k^2 + 89k + 30 = 6k^4 + 39k^3 + 91k^2 + 89k + 30,$$

thereby proving the desired result.

KBS
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Tim Shaw
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1 Answers1

0

Let $S_n=\sum_{i=1}^ni^4$ and define $\tilde S_n=n(n+1)(2n+1)(3n^2 +3n−1)/30$. The goal is to show that $\tilde S_n=S_n$ for all $n\ge1$.

By induction:

  1. Verify that $S_1=\tilde S_1$.
  2. Assume that $S_n=\tilde S_n$ and check whether this implies that $\tilde S_{n+1}=S_{n+1}=S_n+(n+1)^4$.

Enjoy the algebra!

KBS
  • 7,114