says it's   $$\sum_{k=1}^n kz^k = z\frac{1-(n+1)z^n+nz^{n+1}+1}{(1-z)^2}$$  When I tried it on my own I only got as far as:  $$\sum_{i=0}^n ia^i = \frac{n(a^{n+1}-1)}{a-1}-a^n\sum_{i=0}^n(\frac{i}{a^i})$$  But then I realised that the second term is in fact $\sum_{i=0}^n (\frac{1}{a})^ii$ which is only the original problem with $z = \frac{1}{a}$ so I'm a bit stuck.  I should point out that this is a first year undergraduate maths course for engineering. That being said, I'd appreciate if you could point out what course might teach me how to solve this (or a good book).
14153
I want to know the general formula for $\sum_{n=0}^{m}nr^n$ for some constant r and how it is derived.  For example, when r = 2, the formula is given by: $\sum_{n=0}^{m}n2^n = 2(m2^m - 2^m +1)$ according to   Thanks!
32534
I want to find a formula for the sum of this series using its general term. How to do it?  Series $$ S_n = \underbrace{1/3 + 2/21 + 3/91 + 4/273 + \cdots}_{n \text{ terms}} $$  General Term  $$ S_n = \sum_{i=1}^{n} \frac{i}{i^4+i^2+1} $$
8533
Prove that : $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^3}=\frac{\pi^3}{32}.$$    I think this is known (see ), I appreciate any hint or link for the solution (or the full solution).
94432
How would I find the sum below?    $$ \sum_{k=0}^{n} {\frac{1}{k+1} {n \choose k}}=?$$  I need help starting on the problem. Would I first need to utilize the Binomial Theorem proof to get started on it?    Thank you.
86635
I try to solve the following sum:  $$\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k}$$  I'm very curious about the possible approaching ways that lead us to solve it. I'm not experienced with these sums, and any hint, suggestion is very welcome. Thanks.
115573
Let $n \in \mathbb{Z}^+$, prove the identity $$ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=\frac{n(n^{n}-1)}{n+1}$$  First of all  $$ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=n^{n}\Bigg(\sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{-k}}{k+1} \Bigg)$$ $$=n^n\Bigg(\sum_{k=1}^{n-1} \binom {n}{k} \bigg(1-\frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k}\bigg)\Bigg)$$ $$=n^n \sum_{k=1}^{n-1} \binom {n} {k} \frac{1}{n^k}-n^n \sum_{k=1}^{n-1} \binom {n}{k}  \bigg( \frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k} \bigg)$$  We have for the first sum $$(1+\frac{1}{x})^n = \sum_{k=0}^n \binom{n}{k}\frac{1}{x^k}.$$ For the second sum  $$(1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k.$$  Integrating both sides from $0$ to $x$, we see that   $$\frac{(1+x)^{n+1}-1}{n+1} = \sum_{k=0}^n \binom{n}{k}\frac{x^{k+1}}{k+1}.$$  Putting $x=1$ yields $$\frac{2^{n+1}-1}{n+1}=\sum_{k=0}^n \binom{n}{k}\frac{1}{k+1}$$  Here where I have stopped. I could not get them similar for what I have. Would someone help me out !
100263
I'm trying to show that      $$\sum_{k=0}^n (-1)^k\frac{{ {n}\choose{k}}}{{ {x+k}\choose{k}}} = \frac{x}{x+n}$$   I've tried expanding this using definition of binomial coefficient, then binomial expansion, now I'm using induction to simplify $\sum_{k=0}^m (-1)^k\frac{{ {n}\choose{k}}}{{ {x+k}\choose{k}}}$, but it looks very inelegant, is there a better way?
9
I am stuck with the identity  $$ F_{2n} = \sum_{k=1}^n \binom{n}{k} F_k, $$  which happens to be formula . I am using induction, but so far without too much result.  $$ \sum_{k=1}^{n+1} \binom{n+1}{k} F_k = \\ F_{n+1} + \sum_{k=1}^n \binom{n+1}{k} F_k = \\ F_{n+1} + \sum_{k=1}^n \binom{n}{k} F_k + \sum_{k=1}^n \binom{n}{k-1}F_k =\\ F_{n+1} + F_{2n} + \sum_{k=1}^n \binom{n}{k-1}F_k, $$   I am stuck with the sum $ \sum_{k=1}^n \binom{n}{k-1}F_k$, because, on the other hand   $$ F_{n+1}+ \sum_{k=1}^n \binom{n}{k-1}F_k = F_{2n+1} $$  Which would conclude the proof. What am I missing?
2819
$$\sum_{k=2}^{\infty}\frac{1}{k^2-1}$$   I found that:  $$\sum_{k=2}^{\infty}\frac{1}{k^2-1}=\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$$  and $\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$ is a telescopic series so we need $lim_{n \to \infty} \frac{1}{2}-\frac{1}{2(n+1)}=\frac{1}{2}$ but the answer is $\frac{3}{4}$
38306
How do you find an algebraic formula for $\sum_{k=1}^n k^3$? I am able to find one for $\sum_{k=1}^n k^2$, but not $k^3$. Any hints would be appreciated.
67178
I'm wondering about how to show compute this series: $$\sum_{n=0}^{\infty}(-1)^{n-1}\binom{1/2}{n}$$ My approach was to use the general formula of the binomial series, which is: $$(1+z)^r=\sum_{k=0}^{+\infty}z^{k}\binom{r}{k}$$ Yet this can't be used because in this case, we have $|z|=1$. Thus, is there any method that I can use for this?
113139
I was trying to find a formula for the series   $$\sum_{k=1}^n \frac{1}{k(k+1)(k+2)} =? $$  I tried to break this into partial fractions...To see if I could telescope this series.. The partial fraction went like this  $$\frac{1}{k(k+1)(k+2)}=\frac{1}{2k}-\frac{1}{k+1}+\frac{1}{2k+4}$$  But the terms are so random that they hardly cancel.... I also tried partial sums but couldn't make much headway....Any help to solve this would be appreciated
33804
$$\sum_{r=1}^nr(r+1)=\frac{n(n+1)(n+2)}{3}$$  could you help me with how exactly I work this out?
120733
More specifically, I'm supposed to compute $\displaystyle\sum_{k=1}^{n} \frac 1 {k(k + 1)} $ by using the equality $\frac 1 {k(k + 1)} = \frac 1 k - \frac 1 {k + 1}$ and the problem before which just says that, $\displaystyle\sum_{j=1}^{n} a_j - a_{j - 1} = a_n - a_0$.  I can add up the sum for any $n$ but I'm not sure what they mean by "compute".   Thanks!
54720
Mathematica tells me that  $$\sum _{k=0}^n { n \choose k} \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}.$$  Although I have not been able to come up with a proof.  Proofs, hints, or references are all welcome.
68613
$\sum_{n=0}^{\infty}z^{n!}$  My textbook says its answer is 1. But I think the sequence is not power series because it cannot be expressed as the form of $CnZ^n$.   Am I right? Or where do I have a mistake?
37643
$$\sum_{k=0}^nq^k = \frac{1-q^{n+1}}{1-q}$$  I want to prove this by induction. Here's what I have.  $$\frac{1-q^{n+1}}{1-q} + q^{n+1} = \frac{1-q^{n+1}+q^{n+1}(1-q)}{1-q}$$  I wanted to factor a $q^{n+1}$ out of the second expression but that 1- is screwing it up...