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Prove by induction for $d,n \in \mathbb{Z}^+$ that $$ \sum_{k=0}^{n}d^{k}\geq \frac{(n+1)^{d+1}}{d+1}. $$

The base case for $n = 1$ makes $1 \geq 1$ which passes. Then I found that $$ \sum_{k=0}^{n}d^{k} \cdot d^{k+1} \geq \frac{(n+2)^{d+1}}{d+1}. $$ Now, here I am lost, I know that you can replace $$\sum_{k=0}^{n}d^{k}$$ with the right side of the original inequality, but I don't know how to use that to my advantage.

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  • $\begingroup$ it is not correct for $d=1$: $n+1 \geq \frac{(n+1)^2}{2}$ $\endgroup$
    – corindo
    Commented Jan 19, 2015 at 4:48

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For $n=1$, LHS $=1+d$, RHS $=\frac{2^{d+1}}{d+1}$. It's not true that $1+d\ge\frac{2^{d+1}}{d+1}$ for all $d$.

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