What kind of formula is there that can be used for calculating the sum of power of $x$ of numbers from $1$ to $a$?  I know that the sum of numbers from $1$ to $a$ is $\ (n^2 + n)/ 2  \ $ and that the sum of the squared numbers from $1$  to $a$ is $n^3/3+n^2/2+n/6 $.  If possible, please give a solution that excludes Bernoulli numbers and works for negative numbers too.
26839
I have been wanting to understand how to find the sum of this series.   $$1^p + 2^p + 3^p +{\dots} + n^p$$  I am familiar with Gauss' diagonalised adding trick for the sum of the first $n$ natural numbers.  I can prove the formulas for  $$\begin{align}   \sum_{1}^{n} k^2 &amp;= \frac{n(n+1)(2n+1)}{6}\\  \sum_{1}^{n} k^3 &amp;= \frac{n^2(n+1)^2}{4}\\ \sum_{1}^{n} k^4 &amp;= \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} \\ \end{align}$$  With mathematical induction. But, beyond that even proofs with mathematical induction are difficult.  I'm interested in learning the theory and the proof behind Faulhaber's formula. What is the knowledge required to understand this proof ?
45648
Since everyone freaked out, I made the variables are the same. $$ \sum_{x=1}^{n} 2^{x-1} $$  I've been trying to find this for a while. I tried the usually geometric equation  but I couldn't get it right (if you need me to post my work I will). Here's the outputs I need:  1, 3, 7, 15, 31, 63  If my math is correct.
13550
I don't understand at all why:  \begin{equation} \sum\limits_{n=0}^\infty p_n x^n = \prod\limits_{k=1}^\infty (1-x^k)^{-1} \end{equation}  Where $p_n$ is the number of partitions of $n$. Specifically how can the sum of positive numbers ever be factored into a fraction or factors to the $-1$ power? Is this something purely notational or am I missing something big?  Thanks for all the help!
19671
Calculate $$\sum_{k=1}^{\infty}\frac{1}{2^{k+1}-1}.$$    I used Wolfram|Alpha to compute it and got it to be approximately equal to $0.6$. How to compute it? Can someone give me a hint or a suggestion to do it?
26130
Why is $\sum_{k=0}^n (_k^n) 2^k$ simplified as $3^n$ and not just $2^n$ ? The answer uses the binomial theorem as a solution: let $a=1$ and let $b=2$. So $1+2 = 3$. But since $1+2 = 3$ why do we not use $1+1 = 2$?
106527
I believe I see that $a_n = 2^n(a_0+1) - 1$ but am somewhat uncertain where to proceed afterwards. I am a complete beginner at number theory and would appreciate it if someone could point me in the right direction--surely there is some obvious argument I am missing.
60398
I'm wondering if there is a formula for the value of $(1+i)^n+(1-i)^n$?  I calculated the first terms starting with $n=1$ to be, in order, $2$, $0$, $-4$, $-8$, $-8$, $0$, $16$, $\dots$  So it seems to be some sequence of positive and negative powers of $2$ with $0$s thrown in. Is there a more explicit formulation of what $(1+i)^n+(1-i)^n$ is, based on $n$?   With the binomial theorem, I get it equal to $$ \sum_{k=0}^n\binom{n}{k}i^{n-k}(1+(-1)^{n-k}). $$ Can this be made nicer? Thanks.
31453
I'm well aware of the combinatorial variant of the proof, i.e. noting that each formula is a different representation for the number of subsets of a set of $n$ elements. I'm curious if there's a series of algebraic manipulations that can lead from $\sum\limits_{i=0}^n \binom{n}{i}$ to $2^n$.
40271
I've got the following sum which I'm trying to figure out:  $$\sum_{x=0}^n 2^{-x}$$  Wolframalpha tells me that it's equal to $2 - 2^{-n}$ but I am interested in figuring out why, and how to get that result by hand.  Any help is greatly appreciated. Thanks.  Also, as you can tell, I could really use some help with MathJax...
27799
Can someone tell me how to get a closed form for  $$\sum_{k=1}^n k^p$$  For $p = 1$, it's just the classic $\frac{n(n+1)}2$.   What is it for $p &gt; 1$?
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?
71552
I need to find a clear formula (without summation) for the following sum:  $$\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$$  Well, the first few elements look like this:  $1,1,1,2,2,2,2,2,3,3,3,...$  In general, we have $(2^2-1^2)$ $1$'s, $(3^2-2^2)$ $2$'s etc.  Still I have absolutely no idea how to generalize it for $n$ first terms...
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.
97715
My Prof wrote that   $\sum_{n=2}^{\infty} n(n-1)x^{(n-2)} = \frac{2}{(1-x)^3}$  when $-1&lt; x&lt;1$  I do not understand where the above formula comes from. Can someone please explain the steps or provide a hint?   Thanks!
46535
find $$\sum_{k=0}^{t}(-1)^k\binom{t}{k}^2$$ for $t=2n$ then for $t=2n+1$  I tried by expand $(1-x)^n(1-x)^n$, with no result.  Any Help ?
80878
I need a hand in showing that $$ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$$   Thanks in advance for any help.
102474
I need help with finding the formula and proving it by induction. Am stuck, but the professor says we should know this by now.     Find the formula for $1 + \sum \limits _{j=1} ^n j! j$, and show work proving the formula is correct using induction.