<s>[INST] Find sum of first $n$ terms of the series : $1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+\dots$ The main question is:

Find sum of first $n$ terms of the series : $1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+\dots$

My approach:
Initially, nothing clicked, so I went forward with simplifying the series.
So, we get, after simplifying,
$S$(The sum of series) = $1+3+6+10+15+21+\dots$
So, I try to write $S$ as,
$1+(4-1)+(9-3)+(16-6)+\dots$
So what I finally get is, 
$S$=$(1+4+9+16+25+\dots) - (1+3+6+10+\dots)$
Therefore,
$$2S=\sum_{n=1}^n{n^2}$$
Thus,
$S$ = $\frac{n(n+1)(2n+1)}{12}$
But, the answer given in my textbook is 
$\frac{n(n+1)(n+2)}{6}$
My answer is not matching. Please help me by pointing out my mistake or providing a new approach and solution altogether.
 [/INST] Recall that
$$x_n=\sum_{k=1}^n k=\frac{n(n+1)}{2}$$
and 
$$y_n=\sum_{k=1}^n k^3=\frac{n^2(n+1)^2}{4}$$
so 
$$\frac{y_n}{x_n}=x_n$$
so you need to recall the sum
$$\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$$
to compute the desired sum $\sum_{k=1}^n \frac{y_k}{x_k}$.
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