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I have this induction problem I’ve been trying to solve. I can’t seem to close on it given my approach. Maybe my approach is all wrong, I’m not certain. Here is what I’ve done with it so far.

Prove by induction: $1^³+2^³+3^³+···+n^³ < \frac{1}{2}n^4$ ; $∀n∈ℕ$, $n≥3$

Proof:

  1. Base step: when $n=3, 3³=27<\frac{1}{2}3^4 =40.1$

  2. Induction step: Assume $n=k$ is true. For $n=k+1$,

$1³+2³+3³+···+(k+1)³ < \frac{1}{2}(k+1)^4$

⇒ $1³+···+k³+3k²+3k+1< \frac{1}{2}(k^⁴ + 4k^³+6k+4k+1)$

Distribute the $\frac{1}{2}$ on the RHS:

⇒ $1^³+···+k^³+3k^²+3k+1< \frac{1}{2} k⁴ + 2k³+3k²+2k+\frac{1}{2}$

Subtract $3k^²+2k$ from the LHS and RHS:

⇒ $1^3+···+k^3+k+1 < \frac{1}{2}k^4 + 2k^3+\frac{1}{2}$

Subtract $1$ from the LHS and $\frac{1}{2}$ from the RHS and rewrite RHS

⇒ $ 1^³+···+k^³+k < \frac{1}{2}k^4 + 2k^3 = \frac{1}{2}k^4 + k^3+ k^3$

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  • $\begingroup$ There is manipulation, not all correct. There is no use made of the induction hypothesis. You need to show that $\frac{k^4}{2}+(k+1)^3\le \frac{(k+1)^4}{2}$. $\endgroup$
    – André Nicolas
    Sep 29, 2015 at 21:29
  • $\begingroup$ You are starting your induction step by assuming what you want to prove, and you don't want to do that. $\endgroup$
    – user84413
    Sep 29, 2015 at 21:40

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