Finding $\sum_{k=1}^na_k$ knowing that $a_1 + 2a_2 + ... + ka_k = \frac{k+1}{k+2}$. I was just solving a competitive programming question, wherein I found out that a formula can be used for solving it efficiently.
Problem statement: http://www.spoj.com/problems/TOHU/ I tried a lot to derive a formula for finding the value of Sn (given the value of n) directly, but to no avail. Any idea about how this is to be done?
The problem statement is as follows: 
Tohu wants to enter Moscow State University, and he must pass the math exam. He does not know math, and asks you to help him. 
The problem is to find the sum 
$$S_n = a_1 + a_2 + ... + a_n$$ of the sequence $\{a_n\}$ on condition for all  $$k \in \mathbb{N}\ \colon a_1 + 2a_2 + ... + ka_k = \frac{k+1}{k+2}.$$
 A: You may observe that
$$
ka_k=(ka_k+...+2a_2+a_1)-((k-1)a_{k-1}+...+2a_2+a_1)
$$ giving
$$
ka_k=\frac{k+1}{k+2}-\frac{k}{k+1}
$$ that is
$$
a_k=\frac12\left(\frac1k-\frac{1}{k+1}\right)-\frac12\left(\frac{1}{k+1}-\frac{1}{k+2}\right)
$$ Then summing, terms telescope and we easily get

$$
S_n=\sum_{k=1}^na_k=\frac{1}{4}-\frac{1}{2 (1+n)}+\frac{1}{2 (2+n)}.
$$

A: $$a_1+2a_2+...+ia_i={i+1\over i+2}$$
$$a_1+2a_2+...+({i-1})a_{i-1}={i\over i+1}$$
Subtracting these two gives 
$$ia_i={i+1\over i+2}-{i\over i+1}={1\over (i+1)(i+2)}$$
So the desired series is $$\sum_{i=1}^{n} a_i=\sum_{i=1}^{n} {1\over i(i+1)(i+2)}$$
which is a telescopic series.
A: Let:
$$ T_k = \sum_{j=1}^{k} j a_j = \frac{k+1}{k+2}.\tag{1} $$
We have:
$$ T_{k+1}-T_{k} = (k+1) a_{k+1} = \frac{1}{(k+2)(k+3)}\tag{2}$$
from which it follows that:
$$ a_k = \frac{1}{k(k+1)(k+2)} = \frac{\frac{1}{2}}{k}-\frac{1}{k+1}+\frac{\frac{1}{2}}{k+2}\tag{3}$$
and, by telescopic property,

$$ S_k = \sum_{j=1}^{k} a_j = \frac{k(k+3)}{4(k+1)(k+2)}.\tag{4}$$

