I have a question that I don't know how to solve.     For which $n \in \mathbb{N}$ do we have $$\sum\limits^n_{k=0} k^3 = (\sum\limits^n_{k=0} k)^2\quad ?$$.   Here's what I've tried:  \begin{align*}\sum\limits^n_{k=0} k^3 = (\sum\limits^n_{k=0} k)^2 &amp; \Leftrightarrow 1+8+27+\cdots+n^3 = (1+2+3+\cdots+n)^2\\ &amp; \Leftrightarrow \sqrt{1+8+27+\cdots+n^3} = 1+2+3+\cdots+n \end{align*}
11043
I recently proved that  $$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$$  using mathematical induction.  I'm interested if there's an intuitive explanation, or even a combinatorial interpretation of this property. I would also like to see any other proofs.
45758
I'm studying for a midterm and need some help with proving summation     $$\sum\limits_{k=0}^n\binom{n}{k}2^k = 3^n$$   using the binomial theorem.  This is what I've been thinking so far:   In the binomial theorem, we set $x = 0$ and $y = 2$, so:  $3^n = (x+y)^n = \sum\limits_{k=0}^n\binom{n}{k}y^k$  $ = \sum\limits_{k=0}^n\binom{n}{k}2^k$  Am I getting this correct so far or completely wrong? Any help would be appreciated. Thanks.
27537
closed form for  $$\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+...+\binom{n}{n}$$  I tried to solve it by :   $$\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+...+\binom{n}{n}=\sum_{k=0}^{n/3}\binom{n}{3k}$$  $$\sum_{k=0}^{n/3}\binom{n}{3k}=\sum_{k=0}^{n/3}\frac{1}{2\pi i}\int_{\left |z  \right |=1}\frac{(1+z)^n}{z^{3k+1}}dz$$  then using geometric series but i got no result .  what is your suggest to solve ?
31109
$$S=\sum_{k=1}^\infty\left(\frac{k}{k+1}\right)^{k^2};\hspace{10pt}S_n=\sum_{k=1}^n\left(\frac{k}{k+1}\right)^{k^2}$$ Let $S_n$ represent its partial sums and let $S$ represent its value. Prove that $S$ is finite and find an n so large that $S_n$ approximate $S$ to three decimal places.   Solution: first of all, I think that we Will use L'hopital rule and then use root test while starting to solve this. But how?
106328
Is there a closed form for the following?   $$\sum_{k=0}^{m}  {\binom {m}{k}} a^{k} (b+ck)^N$$      how about a pretty limit for large $b$.   I have tried using the binomial expansion for the $(b+ck)^N$ to convert the above expression to double sum and then tried to simplify the double sum which was not successful. Any idea if I have find a closed form for it?
4283
I'm looking for a formula to calculate the sum of $(a+b+c+d+...)^n$ but with coefficients equal to 1. For example in $(a+b+c)^2$. I want the sum of $a^2 + b^2 + c^2 + ab + bc + ca$. And for $(a+b+c+d)^3$, I want the sum of $a^3 + b^3 + c^3 + a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + abc$. Similarly, I want the sum of expansion of $(a+b+c+d+e+....)^n$ with coefficient equal to 1. I tried to find the pattern by expanding the $(a+b+c)^2$. And I found that its formula is $a^2 + b^2 + c^2$ + $\frac{((a+b+c)^2 - a^2-b^2-c^2)}{2}$. This works for the power of 2. But fails on other powers. Any help will be really appreciated.
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$\sum\limits_{n=1}^{\infty}\frac{1}{1+z^n}$, $|z|&gt;1$.  There are two facts that my professor uses that I am confused about.  The first is: $|1+z^n| \geq ||z|^n-1|$, I believe this is true for any $|z|$.  The other is: $\frac{1}{|z|^n-1} \leq \frac{2}{|z|^n}$, I believe this is also true for any $|z|$.  Can anyone prove these statements for me?
68775
Who knows about the summation of this series: $$\sum\limits_{i=1}^{n}i^m $$  where $m$ is constant and $m\in \mathbb{N}$?  thanks
19197
Can you please show me how to get from the left side to the right side?  $$\sum\limits_{k=0}^{20}\binom{50}{k}\binom{50}{20-k} = \binom{100}{20}$$
10614
Show that $$\sum_{k=0}^n(-1)^k\binom nk=0$$ So for odd $n$ we have an even number of terms. So $\binom nk=\binom n{n-k}$ which have opposite signs. Thus the sum is 0.  For even $n$ we have that $$\sum_{k=0}^n(-1)^k\binom nk= \binom n0+\sum_{k=1}^{n-1}(-1)^k\binom nk+\binom nn$$ Now $$\sum_{k=1}^{n-1}(-1)^k\binom nk= \sum_{k=1}^{n-1}(-1)^k\left[\binom{n-1}k+\binom{n-1}{k-1}\right]$$ What would that sum be in the square brackets?
56468
Firstly, this is a homework problem so please do not just give an answer away. Hints and suggestions are really all I'm looking for.  I must prove the following using mathematical induction:     For all $n\in\mathbb{Z^+}$, $1+2+2^2+2^3+\cdots+2^n=2^{n+1}-1$.   This is what I have in my proof so far:     Proof: Let $p\left(n\right)$ be $\sum_{j=0}^n 2^j=1+2+2^2+\cdots+2^n=2^{n+1}-1$. Then $P\left(0\right)$ is    \begin{align} \sum\limits_{j=0}^0 2^j=1=2^1-1.\tag{1} \end{align} Hence    $P\left(0\right)$ is true. Next, for some $k\in\mathbb{Z^+}$, assume    $P\left(k\right)$ is true. Then \begin{align}  \sum\limits_{j=0}^{k+1}2^j&amp;=\left(\sum\limits_{j=0}^k  2^j\right)+2^{k+1}\tag{2}\\ &amp;=2^{k+1}-1+2^{k+1}\tag{3} \\ &amp;=2\cdot 2^{k+1}-1\tag{4}\\ &amp;=2^{k+2}-1\tag{5}\\ &amp;=2^{\left(k+1\right)+1}-1.\tag{6} \end{align}   Therefore, $P\left(n\right)$ holds for all $n$.$\;\;\;\;\blacksquare$   Is everything right? My professor went over the methodology in class quite fast (and I had a bit too much coffee that night) and I want to make sure I have it down for tomorrow night's exam.
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.......
7372
I'm not sure if I should call this "the sum of a sum" but I can't think of another phrase for it.   Anyhow, I'm confused about excercises such as this one  $$\sum_{j=2}^{\infty} \left( \sum_{k=2}^{\infty} k^{-j} \right)$$  where I want to compute the sum. I'm thinking maybe I could rewrite it as  $$\sum_{k=2}^{\infty} \frac{1}{k^{2}} + \sum_{k=2}^{\infty} \frac{1}{k^{3}}+... $$  but I don't know where to go from there or if that is even correct thinking. (I just began my first course in linear analysis.)
123943
$$n!=\sum_{k=0}^n \binom{n}{k}(n-k+1)^n(-1)^k$$  Could anyone give the proof of the above equation? Thanks in advance!
122819
I've got the following finite sum: $s_{n}=\sum\limits_{k=0}^{n}\binom{n}{k}^2p^k$ (esp. if $p$ is a function of $n$, like $p=\frac1{n}$), which can be rewritten as $s_{n}=\sum\limits_{k=0}^{n}\binom{n}{k}\sqrt p^k\binom{n}{k}\sqrt p^k$. Using the generating function approach (from Graham, Knuth and Patashnik) with each of the polynomials in the sum, I get the expression $(1+\sqrt p x)^{2n}$ and the coefficient at the n-th term $x$ turns out to be $\binom{2n}{n} \sqrt p^n$. But comparing this result to the computational value, it turns out to be incorrect. Where did I make a mistake? Are there certain limitations on the generating function method?
55711
Why is $\displaystyle\sum\limits_{k=0}^{n}(-1)^k\binom{n}{k}^2=(-1)^{n/2}\binom{n}{n/2}$ if $n$ is even ?   The case if $n$ is odd, is clear, since $\displaystyle(-1)^k\binom{n}{k}+(-1)^{n-k}\binom{n}{n-k}=0$ (we have $(n+1)/2$ such pairs.)  but if $n$ is even we have no symmetry, I tried to consider the odd and even terms separately but with no success.   Do you have an idea ? Thanks in advance.
50671
I was trying to solve a problem and got stuck at the following step:  Suppose ${n \to \infty}$ .  $$\lim \limits_{n \to \infty} \frac{n^3}{n^3} = 1$$  Let us rewrite $n^3=n \cdot n^2$ as $n^2 + n^2 + n^2 + n^2 \dots +n^2$,$\space$ n times.  Now we have  $$\lim \limits_{n \to \infty} \frac{n^3}{n^3} = \frac {n^2 + n^2 + n^2 + n^2 + n^2 \dots +n^2}{n^3} $$  As far as I understand, we can always rewrite the limit of a sum as the sum of limits ...  $$\dots = \lim \limits_{n \to \infty} \left(\frac{n^2}{n^3} + \frac{n^2}{n^3} + \dots + \frac{n^2}{n^3}\right)$$  ...but we can only let ${n \to \infty}$ and calculate the limit if all of the individual limits are of defined form (is this correct?). That would be the case here, so we have:  $= \dots \lim \limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n}\right) =$[ letting ${n \to \infty}]$ $= 0 + 0 + \dots + 0 = 0$  and the results we get are not the same.  Where did I go wrong?