Q: Calculate     $$S=1+{3\over4}+{3\times5\over4\times8}+{3\times5\times7\over4\times8\times12}+\dots$$   I calculated that the $n$th term is given by $$ \begin{align} A_n&amp;={3\times5\times7\times\dots\times(2n+1)\over4\times8\times12\times\dots\times4n}\\ &amp;={{(2n+1)!\over2\times4\times8\dots\times2n}\over4^nn!}\\ &amp;={(2n+1)!\over8^n(n!)^2} \end{align}$$ Hence $$S=\sum_{n=0}^{\infty}{(2n+1)!\over8^n(n!)^2} =\sum_{n=0}^{\infty}{\binom{2n}{n}(2n+1){1\over8^n}}$$  I'm not sure how to proceed  I feel like generating function for $\sum{\binom{2n}{n}(2n+1)}$ would be helpful but don't know how to derieve it.
6914
How to find infinite sum How to find infinite sum $$1+\dfrac13+\dfrac{1\cdot3}{3\cdot6}+\dfrac{1\cdot3\cdot5}{3\cdot6\cdot9}+\dfrac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $$   I can see that 3 cancels out after 1/3, but what next? I can't go further.
60124
It's asked to simplify the sum  $$\displaystyle\sum_{0\le i &lt;j\le n+1} \binom{n}{i} \times \binom{n+1}{j} $$  inspecting first values of $n$ shows the sum if apparently equal to $4^n$  I tried re-writing the sum as   $$\displaystyle\sum_{j=1}^{n+1} \binom{n+1}{j} \displaystyle\sum_{i=0}^{j-1} \binom{n}{i}  $$ but that doesn't seem to lead us to the result.  Any suggestions are welcome.  Thanks.
119758
Is it true that $\sum_kk\binom{n}{k}^2=n\binom{2n-1}{n-1}$? (I proved it using generating functions). Could you help me to prove it combinatorially? please
21165
How to simplify this best  $(a_1 + a_2 +a_3+... +a_n)^m$ for  $m=n, m&lt;n, m&gt;n$  I could only get  $\sum_{i=0}^{m}\binom{m}{i}a_i^i\sum_{j=0}^{m-i}\binom{m-i}{j}a_j ... $
73531
If you expand $(x_1+x_2+\cdots+x_k)^n$, how many terms will there be once you collect terms with equal monomials? What is the sum of all the coefficients?  I'm kind of lost here. This came up with other questions on the topic of binomial coefficients, and i have figured those out. I have spent almost an hour trying to figure where to start and haven't made any progress.
61469
I've tried this but didn't get the answer : Let $S=2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$  Using $n^3-(n-1)^3 = 3n^2-3n+1$, \begin{align} S &amp;= 3(2014^2)-3(2014)+1+3(2012^2)-3(2012)+1+\ldots+3(2^2)-3(2)+1 \\&amp;= 3\left ( 2014(2013)+2012(2011)+2010(2009)+ \ldots+2(1)  \right ) + 1(1007) \\&amp;= 3\left ( \sum_{n=1}^{1007}2n(2n-1) \right )+1007\\ =&amp; \left ( \sum_{n=1}^{1007}4n^2-\sum_{n=1}^{1007}2n \right )+1007 \\=&amp;\frac{12(1007)(1008)(2015)}{6}-\frac{2(1007)(1008)(3)}{2}+1007 \end{align} This is divisible by $1007$ but not by $1007^2$ which is the correct answer. Where have I gone wrong ?
33410
I have to prove that: $$ \binom n 0 ^2  + \binom n 1 ^2  + \dots + \binom n n ^2 = \binom { 2n} n $$  I don't want a complete solution, but only a hint.
52365
Background: This is a step from a longer proof/exercise that  $\sum_{n=1}^{\infty}\tau (n)^2/n^s=\zeta(s)^4/\zeta(2s)$ for $\sigma&gt;1$  Expanding the sum and using counting I get:   $\frac{1+t}{(1-t)^3}=(1+t)(1+{3\choose 1}t+\left ({3\choose 1}+{3\choose 2}\right )t+\left({3 \choose1}+3!+1\right )t^2 \ldots=\sum_{n=0}^{\infty}(1+n)^2t^n$$  So the final coefficients are each coefficient in the infinite sum plus the coefficient before it because of the $(1+t)$ This gives 1, 4, 9, 16 How would one calculate explicitly (as the author likely did) that the coefficients of $\frac{1+t}{(1-t)^3}$ are $(n+1)^2$ rather than counting and recognizing a pattern?
41206
$$\frac{1}{1 \times3} + \frac{1}{2\times5}+\frac{1}{3\times7} + \frac{1}{4\times9}+\cdots $$  How to find sum of this series?  I tried this: its $n$th term will be = $\frac{1}{n}-\frac{2}{2n+1}$; after that  I am not able to solve this.
32534
I want to find a formula for the sum of this series using its general term. How to do it?  Series $$ 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} $$
53213
i know the summation of $1 \times 2 + 2\times 3 + 3\times 4 = \frac{k(k+1)(k+2)}3$ but this one has negative terms. so we need to find below  $$\sum_{i=1}^n(-1)^{i+1} (i(i+1))$$  e.g. $1\times 2 - 2\times 3 + 3\times 4 ....$  if you can share the derivation then it would be very great.
48055
Student here! Just reading Liebecks Introduction to pure mathematics for fun and I made an attempt at proving the total number of subsets of S is equal to $2^n$.   I realized that the total number of subsets of S is just:  $ \binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n} + 1$  And so I put together the cool looking formula  $ \displaystyle \sum_{r=1}^n \frac{n!}{r!(n-r)!} = 2^n - 1$  But I have no clue how to approach proving this formula analytically. Any help?    So if I try a proof by induction:  $\displaystyle \sum_{r=1}^n \frac{n!}{r!(n-r)!} = 2^n - 1$  $\displaystyle \sum_{r=0}^n \frac{n!}{r!(n-r)!} = 2^n$  obviously true for $n = 1$:  $1 + \frac{1}{1!0!} = 2 = 2$  assume true for $n$, I could multiply both sides by $2$?  $\displaystyle 2\sum_{r=0}^n \frac{n!}{r!(n-r)!} = 2^{n+1}$  How could I go about proving that  $\displaystyle 2\sum_{r=0}^n \frac{n!}{r!(n-r)!}  = \sum_{r=0}^{n+1} \frac{(n+1)!}{r!((n+1) - r)!}$
39097
For $n\in\mathbb N$, $$a_{n+1}=2a_n+\frac{1}{a_n},\quad a_1=1. $$ Can any one give an explicit formula for all $a_n$? If such an explicit general formula doesn't exist, please explain it. I've tried to figure out the $n$-iterated function $f^{(n)}$ where $f(x)=2x+1/x$ or even $f(\tan(t))$. But in either cases, I failed.Since the recurrence isn't linear nor homogeneous,the generating function method doesn't apply here.
1462
The binomial sum   $$s_n=\binom{n+1}{0}+\binom{n}{1}+\binom{n-1}{2}+\cdots$$   satisfies the Fibonacci relation.   I failed to prove that $\binom{n-k+1}{k}=\binom{n-k}{k}+\binom{n-k-1}{k}$... Any hints or suggestions?
51481
I used WolframAlpha to calculate a sum but it didn't show me the way :( Anybody has a hint or a solution for proving this sum? $$\sum_{n=0}^N\binom{2N-n}N2^n(n+1)=(1+2N)\binom{2N}N$$
68791
Simplifying this expression$$1\cdot\binom{n}{0}+ 2\cdot\binom{n}{1}+3\cdot\binom{n}{2}+ \cdots+(n+1)\cdot\binom{n}{n}= ?$$ $$\text{Hint: } \binom{n}{k}= \frac{n}{k}\cdot\binom{n-1}{k-1} $$
25151
I need to express $$1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb  + \frac{1}{n + 1}\binom{n}{n}$$ in a simplified form.  So I used the identity $$(1+x)^n=1 + \binom{n}{1}x + \binom{n}{2}x^2 + \dotsb   + \binom{n}{n}x^n$$ Now on integrating both sides and putting $x=1$.  I am getting $$\frac{2^{n+1}}{n+1}$$ is equal to the given expression.But the answer in my book is  $$\frac{2^{n+1}-1}{n+1}.$$ Where does that -1 term in the numerator come from?