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I'm having trouble using Abel's summation formula as $a_1b_1+a_2b_2+...+a_nb_n=(a_1-a_2)(b_1)+(a_2-a_3)(b_1+b_2)+...+(a_{n-1}-a_n)(b_1+b_2+...+b_{n-1})+a_n(b_1+b_2+...+b_n)$

to find the sum of $1+4k+9k^2+...+n^2k^{n-1}$.

I know $1+2k+3k^2+...+nk^{n-1}=\frac{nk^n}{k-1}-\frac{k^n-1}{(k-1)^2}$ from a previous part.

Applying Abel's formula once I get $(-3)(1)+(-5)(1+k)+...-(2n-1)(1+k+k^2+...+k^{n-2})+n^2(1+k+k^2+...+k^{n-1})$

Then I think the next step is to apply it again to get $-2(1)-2(1+1+k)-...-2(1+(1+k)+(1+k+k^2)+...+(1+k+k^2+...+k^{n-3}))+(2n-1)(1+(1+k)+...+(1+k+...+k^{n-2}))+n^2\left(\frac{k^n-1}{k-1}\right)$

$=-2[1+(1+1+k)+(1+1+k+1+k+k^2)+...+(1+(1+k)+...+(1+k+..+k^{n-3}))]+(2n-1)((n-1)+(n-2)k+...+(n-(n-1))k^{n-2})+n^2\left(\frac{k^n-1}{k-1}\right)$

$=-2[1+(1+1+k)+(1+1+k+1+k+k^2)+...+(1+(1+k)+...+(1+k+..+k^{n-3}))]+(2n-1)\left(n\left(\frac{k^{n-1}-1}{k-1}\right)-\frac{(n-1)k^{n-1}}{k-1}+\frac{k^{n-1}-1}{(k-1)^2}\right)+n^2\left(\frac{k^n-1}{k-1}\right)$

$=-2[1+(1+1+k)+(1+1+k+1+k+k^2)+...+(1+(1+k)+...+(1+k+..+k^{n-3}))]+(2n-1)\left(\frac{k^{n-1}-n}{k-1}+\frac{k^{n-1}-1}{(k-1)^2}\right)+n^2\left(\frac{k^n-1}{k-1}\right)$

$=-2[1+(1+1+k)+(1+1+k+1+k+k^2)+...+(1+(1+k)+...+(1+k+..+k^{n-3}))]+\frac{2n-1}{(k-1)^2}\left(k^n-nk+n-1\right)+n^2\left(\frac{k^n-1}{k-1}\right)$

My problem arises when trying to find $[1+(1+1+k)+(1+1+k+1+k+k^2)+...+(1+(1+k)+...+(1+k+..+k^{n-3}))]$, it feels way too contrived and the sums within a sum thing is confusing me, I feel like I'm not taking the right approach, if anyone has a faster way using this formula that would be greatly appreciated.

By the way the solution is $\frac{n^2k^n}{k-1}-\frac{(2n-1)k^n+1}{(k-1)^2}+\frac{2k^n-2}{(k-1)^3}$

Pen and Paper
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  • See https://www.wolframalpha.com/input/?i=sum+n%5E2+z%5E%28n-1%29 – lhf Dec 18 '21 at 11:01
  • @lhf, I know the answer, it is given to me in the answers. I'm asking how to actually do it using Abel's summation formula and not the infinite sum case, just up to $n$ terms. If you know how it would be great to know – Pen and Paper Dec 18 '21 at 11:04
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    not Abel summation, but if you treat $k$ as a real variable then just multiply your previous result by $k$ and then differentiate by $k$ – user8268 Dec 18 '21 at 11:07

2 Answers2

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Hint

In one hand, $$S_n(x)=\sum_{k=1}^nkx^k=x\cdot \frac{d}{dx}\sum_{k=0}^nx^k.$$

Finally, $$\text{Your sum}=\frac{d}{dx}S_n(x).$$

Surb
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Surb’s method is much cleaner, but you can also use the method used to find the sum of geometric series to solve this.

\begin{align} v &= \sum_{k=1}^nk^2x^{k-1} \\ v-vx &= -n^2x^n+\sum_{k=0}^{n-1}(2k+1)x^k \\ &= -n^2x^n+2\sum_{k=0}^{n-1}kx^k+\sum_{k=0}^{n-1}x^k \end{align}

At which point solving for $v$ should be pretty easy (the same method can solve $\sum_{k=0}^{n-1}kx^k$ if you want to use it).

iosce
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