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One is asked to show that $ \sum\limits_{i=1}^{n} k^{p}$ (typo on $i$?) can always be written in the form

$$\frac{n^{p+1}}{p+1}+An^{p}+Bn^{p-1}+Cn^{p-2}+\cdots.$$

The solution states:

The proof is by complete induction on $p$. The statement is true for $p=1$, since $$\sum\limits_{k=1}^{n} k= \frac{n(n+1)}{2}= \frac{n^{2}}{2}+n.$$ Suppose that the statement is true for all natural numbers $\leq p$. The binomial theorem yield the equations $$(k+1)^{(p+1)}-k^{p+1}=(p+1)k^{p}+ \textrm{terms involving lower powers of }k.$$ Adding for $k=1,\dots, n,$ we obtain $$\frac{(n+1)^{p+1}}{p+1}=\sum\limits_{k=1}^{n} k^{p} + \textrm{terms involving} \sum\limits_{k=1}^{n} k^{r} \textrm{ for } r<p.$$ By assumption we can write each $\sum\limits_{k=1}^{n} k^{r}$ as an expression involving powers $n^{s}$ with $s\leq p$. It follows that $$\sum\limits_{k=1}^{n} k^{p}=\frac{(n+1)^{p+1}}{p+1} + \textrm{ terms involving powers of }n \textrm{ less than } p+1.$$

What I tryed based on it and more explicitly:

$$\begin{align}(k+1)^{p+1} &={p+1 \choose p+1}k^{p+1}+{p+1 \choose p}k^{p}+ \cdots +{p+1 \choose 1}k + {p+1 \choose 0}k^{0}\\ (k+1)^{p+1}&= 1\cdot k^{p+1}+{p+1 \choose p}k^{p}+ \cdots +{p+1 \choose 1}k + {p+1 \choose 0}k^{0}\\ (k+1)^{p+1} -k^{p+1} &= (p+1)k^{p}+ \cdots +(p+1)k + 1\cdot k^{0}\\ \end{align}$$

Adding for $k=1,\dots, n,$ is something like

$$\begin{align} {2^{p+1}} -1^{p+1} &= (p+1)1^{p}+ \cdots +(p+1)1+ 1\cdot 1^{0}\\ {3^{p+1}} - {2^{p+1}} &= (p+1)2^{p}+ \cdots +(p+1)2+ 1\cdot 2^{0}\\ & \vdots\\ (n+1)^{p+1} -{n^{p+1}} &= (p+1)n^{p}+ \cdots +(p+1)n+ 1\cdot n^{0}\\ \hline & \hline\\ (n+1)^{p+1} &= (p+1)\sum\limits_{k=1}^{n} k^{p}+ \cdots +(p+1)\sum\limits_{k=1}^{n} k^{1}+ \sum\limits_{k=1}^{n} k^{0} + k^{0}\\ \frac{(n+1)^{p+1}}{(p+1)} &= \sum\limits_{k=1}^{n} k^{p}+ \frac{{p+1 \choose p-1}}{(p+1)}\sum\limits_{k=1}^{n}k^{p-1} + \cdots +\frac{(p+1)}{(p+1)}\sum\limits_{k=1}^{n} k^{1}+ \frac{1}{(p+1)}(\sum\limits_{k=1}^{n} k^{0} + k^{0})\\ \end{align}$$

And assuming the proposition as true for $p-1$ we could write

$$\begin{align} \frac{(n+1)^{p+1}}{(p+1)} &= \sum\limits_{k=1}^{n} k^{p}+ \textrm{terms involving powers of } n\leq p.\\ \sum\limits_{k=1}^{n} k^{p} &= \frac{(n+1)^{p+1}}{(p+1)}+ \textrm{terms involving powers of } n\leq p.\\ \end{align}$$

The question is:

How do $(n+1)^{p+1}$ instead of $n^{p+1}$ in the result still makes the proof valid?

1 Answers1

2

If I understand correctly you are concerned that the last part of the solution gives:

$\sum_{k=1}^{n}k^{p}=\frac{(n+1)^{p+1}}{p+1}+$ terms involving powers of $n\le p$

but the original question asked for:

$\sum_{k=1}^{n}k^{p}=\frac{n^{p+1}}{p+1}+$ terms involving powers of $n\le p$.

But this is not a problem since $(n+1)^{p+1}=n^{p+1}+$ terms involving powers of $n\le p$ by the binomial theorem. You can combine the constants of these new terms with the ones given in the solution. More explicitly, we will have:

$\sum_{k=1}^{n}k^{p}=\frac{(n+1)^{p+1}}{p+1}+\sum_{q=0}^{p}A_{q}n^{q}=\frac{n^{p+1}}{p+1}+\frac{1}{p+1}\sum_{q=0}^{p}\binom{p+1}{q}n^{q}+\sum_{q=0}^{p}A_{q}n^{q}=\frac{n^{p+1}}{p+1}+\sum_{q=0}^{p}\bigg(\frac{\binom{p+1}{q}}{p+1}+A_{q}\bigg)n^{q}$

user71352
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