My maths teacher showed me something on how to calculate sums. Let's take an example:

$$\sum_{k=1}^n k(k+1) = \sum_{k=1}^n k^2 + \sum_{k=1}^n k = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(n+2)}{3} $$

This was an easy one, but I just can't understan how to solve such sums:

$$\sum_{k=1}^n (k-1)k(k+1)\tag{example 1}$$

$$\sum_{k=1}^n \frac{1}{(3n-2)(3n+1)}\tag{example 2}$$

Could anybody help me, please?

I want to understand the idea of solving sums like these, so please, do not be very specific, but help me giving these and maybe some other examples.

  • $\begingroup$ For $\sum k(k+1)$ the teacher maybe noted that $k(k+1)=\frac{1}{2}\left((k+1)^2-k^2 \right)$. Then we get an instant answer for the sum via telescoping. $\endgroup$ – André Nicolas Nov 11 '14 at 18:17
  • $\begingroup$ I don't know if it's the best practice to telescope the sum, is it? $\endgroup$ – Victor Nov 11 '14 at 18:18
  • $\begingroup$ Hard to know what best practice is. It certainly works quickly. $\endgroup$ – André Nicolas Nov 11 '14 at 18:21


The second example is orthogonal to the first, hence a different answer


Set a few values of $n=1,2,3,\cdots,n-2,n-1,n$ to recognize the Telescoping Series

  • $\begingroup$ I think there is something wrong in this formula, because the second fraction would be 1. $\endgroup$ – Victor Nov 11 '14 at 18:17
  • $\begingroup$ @Victor, How much you are getting ? $\endgroup$ – lab bhattacharjee Nov 11 '14 at 18:19
  • $\begingroup$ It's ok right know. $\endgroup$ – Victor Nov 11 '14 at 18:19
  • $\begingroup$ So, given your logic, the sum would be $\frac{1}{3}\sum_{k=1}^n (\frac{1}{3n-2} - \frac{1}{3n+1}) = \frac{1}{3}(1-\frac{1}{3n+1})$, right? $\endgroup$ – Victor Nov 11 '14 at 18:21
  • $\begingroup$ @Victor, I believe the Question should be $$\sum_{k=1}^n\frac1{(3k-2)(3k+1)}$$ Then $$\sum_{k=1}^n\frac1{(3k-2)(3k+1)}=\frac13\left(\frac1{3k-2}-\frac1{3k+1}\right)$$ $\endgroup$ – lab bhattacharjee Nov 11 '14 at 18:23



See Sum of cubes proof and Proving the sum of the first $n$ natural numbers by induction

  • $\begingroup$ See my edit, and please, be patient to explain those things briefly... $\endgroup$ – Victor Nov 11 '14 at 18:06
  • 1
    $\begingroup$ @Victor, Ideally the second example should be a different Question. See my other answer $\endgroup$ – lab bhattacharjee Nov 11 '14 at 18:11
  • $\begingroup$ Should I edit it again? I guess people will become upset :)) $\endgroup$ – Victor Nov 11 '14 at 18:12
  • 1
    $\begingroup$ @Victor, I think, you should better leave it as it is and practise the pattern : One Question, one post in future $\endgroup$ – lab bhattacharjee Nov 11 '14 at 18:14
  • $\begingroup$ Ok. Thank you for your tip $\endgroup$ – Victor Nov 11 '14 at 18:14

You can use the fact that $$ (k-1) k (k+1) = 6\binom {k+1}3 $$ to get $$ \sum_{k=1}^n (k-1) k (k+1)= 6\sum_{k=1}^n \binom {k+1}3 = 6\binom{n+2}4 = \frac{(n+2)(n+1)n(n-1)}{4} $$

For the first sum, you also have $$ \sum_{k=1}^n k(k+1) = 2\sum_{k=1}^n \binom {k+1}2 = 2\binom{n+2}3 = \frac{(n+2)(n+1)n}3 $$

For the second problem, this is a totally different beast. You should write $$ \frac1{(3n−2)(3n+1)} = \frac 13\frac{(3n+1) - (3n-2)}{(3n−2)(3n+1)} = $$

and most terms telescope.


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