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Since I learned easier ways of calculating summations I've been curious as to how I could find formulas for as many equations as possible. I came across the equation $x^2+x$, I've spent quite some time on this problem and could not find a solution. If someone has maybe already done this or have any suggestions on how I could get the formula that would be greatly appreciated.

Example of another summation with a equation: $\sum\limits_{i=1}^n$ = $x^2$

Equation to solve this is $\frac{n(n+1)(2n+1)}{6}$

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    $\begingroup$ You want to find $\sum_{i=1}^N(i^2 + i)$? $\endgroup$ – M.B. Nov 14 '13 at 23:03
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    $\begingroup$ Are you asking about a closed formula for $\sum \limits_{k=0}^n\left(k^2+k\right)$, for every $n\in \Bbb N$? If so, just separate the sum in two well known sums. $\endgroup$ – Git Gud Nov 14 '13 at 23:03
  • $\begingroup$ 1) Yes 3) Yes and thank you Thanks kbball for that edit, I have no idea how that works $\endgroup$ – Harjit Nov 15 '13 at 4:12
  • $\begingroup$ You are using the word "equation" but the mathematical term is "expression" unless you have an $=$ sign. $\endgroup$ – Sammy Black May 19 '14 at 17:40
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Extending what Git Gud said: $$\sum^n_{k=0}\left(k^2+k\right)=\sum^n_{k=0}k^2+\sum^n_{k=0}k=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\frac{n^3}{3}+n^2+\frac{2n}{3}$$

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  • $\begingroup$ So for n=6 we get $\frac{6^3}{3} +6^2 + \frac{6}{6} + \frac{1}{2} = 72 + 36 + 1 + \frac{1}{2} = 109.5?$ $\endgroup$ – Steve ODonnell Nov 14 '13 at 23:55
  • $\begingroup$ When I do the summation of (K^2+k) starting from zero manually I get; 0+2+6+12+20+30+42=112 but using the equation we get 109.5? $\endgroup$ – Harjit Nov 15 '13 at 4:08
  • $\begingroup$ Also wouldn't the equation be (n^3/3+n^2+2n/3)/2. According to the manual summation that makes more sense to me. I am though making an assumption that dividing by 2 is there due to having added two summations, am I right? $\endgroup$ – Harjit Nov 15 '13 at 4:33
  • $\begingroup$ Oops sloppy algebra $\endgroup$ – Ali Caglayan Nov 16 '13 at 2:03
  • $\begingroup$ Oh alright, no worries, thanks! $\endgroup$ – Harjit Nov 16 '13 at 4:00
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Note that $$k^2+k=\frac{1}{3}\left((k+1)^3-k^3-1\right).$$ Thus our sum $\sum_{k=1}^n (k^2+k)$ is equal to $$\frac{1}{3}\left((2^3-1^3-1)+(3^3-2^3-1)+(4^3-3^3-1)+(5^3-4^3-1)+\cdots +((n+1)^3-n^3-1)\right).$$ Observe the nice almost total cancellations. We end up with $$\frac{1}{3}\left((n+1)^3-1^3-n\right).$$

Remarks: $1.$ Since we know $\sum_1^n k$, this gives a way to derive the formula for $\sum_1^n k^2$.

$2.$ The sums $\sum k(k+1)$, $\sum k(k+1)(k+2)$, $\sum k(k+1)(k+2)(k+3)$ and so on are nice, much nicer than $\sum k^2$, $\sum k^3$, $\sum k^4$ and so on.

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  • $\begingroup$ @Ben: Thanks for the fix. $\endgroup$ – André Nicolas Nov 19 '13 at 6:25
  • $\begingroup$ +1 for the telescoping sum explanation, since they lie at the heart of the summation formulas that others are using. $\endgroup$ – Sammy Black May 19 '14 at 17:45
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Note that $$\frac{n^2+n}{2} = {n+1 \choose 2}.$$ Then by induction $$\begin{eqnarray}{n+2 \choose 3}&=&{n+1 \choose 2}+{n+1 \choose 3}\\ &=&{n+1 \choose 2} + \sum_{k=1}^{n-1}{k+1 \choose 2}\\ &=& \sum_{k=1}^{n}{k+1 \choose 2}. \end{eqnarray}$$ So $$\sum_{k=1}^{n}(k^2+k)=2{n+2 \choose 3}.$$ This directly generalizes to binomial sums $$\sum_{k=1}^n{k + m-1 \choose m}.$$ This is particularly obvious if you draw what is going on in Pascal's triangle.

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