# Math Help - Telescopic sums

1. ## Telescopic sums

Using the method of telescopic sums, how do I evaluate the following finite sums?

$S_1 = 3.2.3 + 3.3.4 + ... + 3(n+1)(n+2) + ... + 30906 using {n(n+1)(n+2)}$

$S_2 = 2/(1.2.3) + 2/(2.3.4) + ... + 2/(n(n+1)(n+2)) + ... + 2/(1030200) using {-1/(n(n+1))}$

Thank you

2. Originally Posted by tsal15
Using the method of telescopic sums, how do I evaluate the following finite sums?

$S_1 = 3.2.3 + 3.3.4 + ... + 3(n+1)(n+2) + ... + 30906$ using ${n(n+1)(n+2)}$

$S_2 = 2/(1.2.3) + 2/(2.3.4) + ... + 2/(n(n+1)(n+2)) + ... + 2/(1030200)$ using ${-1/(n(n+1))}$

Thank you
$S_2$ has been asked and replied to in this thread: http://www.mathhelpforum.com/math-he...opic-sums.html. Any further correspondence regarding this sum should be asked there.

3. Originally Posted by mr fantastic
$S_2$ has been asked and replied to in this thread: http://www.mathhelpforum.com/math-he...opic-sums.html. Any further correspeondence regarding this sum should be asked there.
Note that 30906 = 3(101)(102). Therefore

$S_1 = 3 \, [\, 2.3 + 3.4 + .... + (n+1)(n+2) + .... + (101)(102) \, ]$.

Note that $(n + 1)(n+2) = n^2 + 3n + 2$.

Therefore $S_1 = 3 \sum_{n=1}^{100} (n+1)(n+2) = 3 \sum_{n=1}^{100} (n^2 + 3n + 2)$ $= 3 \sum_{n=1}^{100} n^2 + 9 \sum_{n=1}^{100} n + 3 \sum_{n=1}^{100} 2 = 3 \sum_{n=1}^{100} n^2 + 9 \sum_{n=1}^{100} n + 3(2)(100) = \, ....$

You should have formulae you can use for getting $\sum_{n=1}^{100} n$ and $\sum_{n=1}^{100} n^2$.