I am stuck on this problem: Simplify the sum $$ \sum_{k=0}^n {n+k \choose k} 2^{-k}. $$ I multiplied it by $\frac{2^n}{2^n}$ to get $$ \frac{1}{2^n}\sum_{k=0}^n {n+k \choose k} 2^{n-k} = \frac{1}{2^n}\sum_{k=0}^n {n+k \choose n} 2^{n-k} $$ How should I proceed? I tried to use the formula $\sum_{i=r}^n {i \choose r} =  {n+1 \choose r+1}$ but I don't know how to aply it here.
19041
I just found this identity but without any proof, could you just give me an hint how I could prove it? $$2^n = \sum\limits_{k=0}^n 2^{-k} \cdot \binom{n+k}{k}$$  I know that $$2^n = \sum\limits_{k=0}^n \binom{n}{k}$$ but that didn't help me
46049
I want to Calculate $\sum_{k=1}^n k2^{n-k}$. Here's my attempt:  $$\begin{align}  &amp;\sum_{k=1}^n k2^{n-k} = \sum_{k=1}^n k2^{n+1}2^{-k-1} =2^{n+1}\sum_{k=1}^n k2^{-k-1} &amp; &amp;\text{(1)} \\ &amp;k2^{-k-1} = -\frac{d}{dk}2^{-k} &amp; &amp;\text{(2)} \end{align}$$   Plugging $(2)$ to $(1)$,   $$\begin{aligned}  2^{n+1}\sum_{k=1}^n k2^{-k-1}  &amp;= 2^{n+1}\sum_{k=1}^n-\frac{d}{dk}2^{-k} \\  &amp;= -2^{n+1}\sum_{k=1}^n\frac{d}{dk}2^{-k} \\  &amp;= -2^{n+1}\frac{d}{dk}\sum_{k=1}^n2^{-k} \\  &amp;= -2^{n+1}\frac{d}{dk}(1-2^{-n}) \\ &amp;= -2^{n+1}\cdot 0 \\ &amp;= 0 \end{aligned}$$   Obviously, I don't know much about summation and derivative. Can you please let me know where I am doing wrong and give me a hint for this question? Thanks!
70000
I have to prove that:     $$\sum_{k=0}^{n} {{m+k} \choose{m}} = { m+n+1 \choose m+1 }$$   I tried to open up the right side with Pascal's definition that: $$ { n \choose k} = {n-1 \choose {k}} + {n-1 \choose {k-1}}$$  Here is what I came up with, and I am sure it is wrong because it does not equal the left side:  $$ {m+n+1 \choose m+1} = {m+n \choose m+1} + {m+n \choose m} = ... ={m+n \choose m+1} + {m+n-1 \choose m} + {m+n-2 \choose m-1} + ... + {m \choose m+1} = \sum_{k=0}^{n} {m+k \choose m+k-n+1 }  $$  Which, again, probably is wrong because it is not equal $\sum_{k=0}^{n} { m+k \choose m}$. Any help is appreciated
23483
I'm repeating material for test and I came across the example that I can not do. How to calculate this sum: $\displaystyle\sum_{k=0}^{n}{2n\choose 2k}$?
18020
How does one show that $$\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^{n}$$ for each nonnegative integer $n$?  I tried using the  but I guess I am applying it incorrectly. With the snake oil technique we have $$F(x)= \sum_{n=0}^{\infty}\left\{\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}\right\}x^{n}.$$ I think I have to interchage the summation and do something. But I am not quite comfortable in interchanging the summation. Like after interchaging the summation will $$F(x)=\sum_{k=0}^{n}\sum_{n=0}^{\infty}\binom{n+k}{k}\frac{1}{2^k}x^{n}?$$ Even if I continue with this I am unable to get the correct answer.  How does one prove this using the Snake oil technique?  A combinatorial proof is also welcome, as are other kinds of proofs.
86142
I try to simplify to get rid of sum $$ \sum_{k=0}^{n-1}\cos(2 \pi fk)$$  I discover I shall use euler equation to form:  $$ \sum_{k=0}^{n-1}\frac{1}{2}(e^{2 \pi fki}+e^{-2 \pi fki})$$  but how to sum exponentials?
68440
$\displaystyle\sum_{i=0}^n2^i=2^{n+1}-1$.  I don't understand induction so I could use some help.
30637
I have been trying to find the sum $\sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{2n+1}{3^n} $. After some calculation, I got here: $\frac{-6}{8}+\frac{1}{4}+8\sum\frac{k}{9^k}$. I know the result is $\frac{5}{8}$ , and I verified it with Wolfram Alpha.I saw that $\sum_{k=1}^{\infty}\frac{k}{9^k}=\frac{9}{64}$. But I don't know how to prove the last equation: $\sum_{k=1}^{\infty}\frac{k}{9^k}=\frac{9}{64}$. I hope someone could help me or show me another method to find the sum for my initial series.
49569
How to find the sum of this series?  $$\sum_{k=0}^{\infty}\cfrac{{2}^{k}}{\binom{2k+1}{k}}$$  It seems very easy. But I still can not work it out, can anyone help?
30701
$$\sum_{i=1}^n \frac1{n+i} = \sum_{i=1}^{2n}\frac{(-1)^{i+1}}i,\quad \text{ for }n\ge 1.$$  I really have no clue how on to solve this one. It's the last one on my assignment. Would really enjoy some help with this one.
100910
I was stuck while finding the given summation.  $$\sum_{k=1}^rk^2\binom {n-k}{r-k}$$  Since $n$ and $r$ are both constants, so I have first converted the above summation into this:  $$\sum_{k=1}^rk^2\binom {n-k}{n-r}$$  I have no idea how to proceed next. Any help will be appreciated.
49662
How to find the sum $$\sum^n_{k=1} (k^2+k+1)k!$$  What I tried as follows :   $$\sum^n_{k=1} (k^2+k+1)k!$$  =$$\sum^n_{k=1} (k^2)k!+ \sum^n_{k=1} (k)k! + \sum^n_{k=1} (1)k!$$  Now we can write $\sum^n_{k=1} (k^2) $ as $$\frac{n(n+1)(2n+1)}{6}$$ but what to do with $k!$ please guide thanks.......
23389
Proof that $\suṃ̣_{k=1}^{n}k {n \choose k}$ for $n \in \mathbb N$ is equal to $n2^{n-1}$.  As a hint I got that $k {n \choose k} = n {n-1\choose k-1} $.  I tried solving this by induction but, in the inductive step I'm not arriving to the correct result.
56444
Simplify $$\frac{\displaystyle\sum_{k=1}^{2499}\sqrt{10+{\sqrt{50+\sqrt{k}}}}}{\displaystyle\sum_{k=1}^{2499}\sqrt{10-{\sqrt{50+\sqrt{k}}}}}$$  I don't have any good idea. I need your help.
62841
Reading through my textbook I came across the following problem, and I am looking for some help solving it. In the back of the textbook they have shown the final answer but I'm not quite sure how to get there. If I could get some help that would be great, thanks!  Find,  $$\sum_{k=0}^{n}4^k \binom{n}{k}$$  The solution given is,  $$C(7,1)C(4,1)+C(7,2)C(4,2)+C(7,3)C(4,3)+C(7,4)C(4,4)$$
83167
$$2^{n-2} n (n -1) = \sum\limits_{k=2}^n k (k - 1) \binom{n}{k}.$$  I'm completely stumped. I just have no idea how to do this. What I've tried so far has been simplifying the right hand side slightly to $\sum\limits_{k=2}^n \binom{n}{k - 2}$, and then proving that this somehow is equal to some manipulation of the left hand side. I can't seem to find any breakthroughs that way though.  If anyone could help out or point me in the right direction that would be great.  EDIT: I've actually just realized my simplification of the right hand side is wrong. Now I'm even more stuck.
65961
I was trying to simply an expression in an exercise related to randomized algorithms. Here is the expression which I have obtained at the end.  $$ \displaystyle\frac{\displaystyle\sum_{i=1}^{n+k-1} i \binom{n-i}{k-1}}{ \displaystyle{n \choose k}}$$  Is there any way to simplify the numerator so that the whole expression simplifies into a nice closed formula? A combinatorial approach would be greatly appreciated.