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I know that infinite series has been given already for the same series but I haven't seen any finite derivation of it.

I tried to solve it and I come out on $\frac{k - N^2k^{2N+2}}{(1-k)^2}$ but I don't know whether this is correct.

Derivation: \begin{equation} S_N = k + 2k^2 + 3k^3 ... Nk^N \end{equation} \begin{equation} k*S_N = k^2 + 2k^3 + ... (N-1)k^N + Nk^{N+1} \end{equation} Results in: \begin{equation} (1-k)*S_N = k + k^2 + k^3 ... k^N - Nk^{N+1} \end{equation} So \begin{equation} (1-k)*S_N = \frac{k}{1-k} - Nk^{N+1} \end{equation} So $S_N$ is \begin{equation} S_N = \frac{k}{(1-k)^2} - \frac{Nk^{N+1}}{1-k}= \frac{k - N^2k^{2N+2}}{(1-k)^2} \end{equation}

alexSDE
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  • $k + k^2 + k^3+ \dots + k^N$ is not $\frac{k}{1-k}$, use formula for finite sum of geometric series, not infinite – Sil Oct 09 '21 at 18:55
  • It also looks like you made a simplification error in the last line when combining the fractions. But you are on the correct track. – Alan Abraham Oct 09 '21 at 18:56
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    It's traditional to use $k$ or $n$ for integers and $x$ for continuous variables, so most would write this as $\sum_{k=0}^N kx^k$ (although the other way is not incorrect.) Cheers – Jair Taylor Oct 09 '21 at 19:00
  • Not an immediate answer to your question, but if you're interested in solving sums like this you might look into finite calculus. You can read a great intro here, and a wealth of knowledge in Graham, Knuth, and Patashnik's Concrete Mathematics. The particular technique for solving a sum like this is summation by parts, the finite analogue of integration by parts. – HallaSurvivor Oct 09 '21 at 19:01
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    Will do @HallaSurvivor thank you all for the help! – alexSDE Oct 09 '21 at 19:06

2 Answers2

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$$\sum_{k=1}^nkx^k=\sum_{k=1}^\infty kx^k-\sum_{k=n+1}^\infty kx^k.$$ $$\sum_{k=n}^\infty kx^k=x\sum_{k=n}^\infty kx^{k-1}=x\frac d{dx}\sum_{k=n}^\infty x^k=x\frac d{dx}\frac{x^n}{1-x}=\frac{n(1-x)x^n+x^{n+1}}{(1-x)^2}.$$ $$\sum_{k=1}^nkx^k=\sum_{k=1}^\infty kx^k-\sum_{k=n+1}^\infty kx^k=\frac x{(1-x)^2}-\frac{(n+1)(1-x)x^{n+1}+x^{n+2}}{(1-x)^2}=\boxed{\frac{x-(n+1)x^{n+1}+nx^{n+2}}{(1-x)^2}}.$$ In your notation: $$\sum_{x=1}^Nxk^x=\boxed{\frac{k-(N+1)k^{N+1}+Nk^{N+2}}{(1-k)^2}}.$$

bof
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Your general method is on the correct track, but you have made some mistakes. Another method is to consider the function $$f(k)=\sum_{x=0}^n k^x=\frac{k^{n+1}-1}{k-1}$$ Taking the derivative wrt $k$ and multiplying by $k$ on all sides gives $$kf'(k)=\sum_{x=0}^n xk^x=k\cdot \frac{(n+1)k^n(k-1)-(k^{n+1}-1)}{(k-1)^2}$$ $$kf'(k)=\sum_{x=0}^n xk^x=\frac{(n+1)k^{n+2}-(n+1)k^{n+1}-k^{n+2}+k}{(k-1)^2}$$ $$kf'(k)=\sum_{x=0}^n xk^x=\frac{nk^{n+2}-(n+1)k^{n+1}+k}{(k-1)^2}$$ Clearly $\sum_{x=0}^n xk^x=\sum_{x=1}^n xk^x$, so we have $$\boxed{\sum_{x=1}^n xk^x=\frac{nk^{n+2}-(n+1)k^{n+1}+k}{(k-1)^2}}$$

Alan Abraham
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