Help simplifying the expression $\bigg(-1+{n+1 \choose 2}\bigg)+{n+1 \choose 2}+\bigg(1+{n+1 \choose 2}\bigg)+\cdots+\bigg(n-2+{n+1 \choose 2}\bigg)$

Question

I need help simplifying the following sum

If $n \geq 2$ then simplify in terms of $n$ - $$s_n =\bigg(-1+{n+1 \choose 2}\bigg)+{n+1 \choose 2}+\bigg(1+{n+1 \choose 2}\bigg)+\cdots+\bigg(n-2+{n+1 \choose 2}\bigg)$$

Here is my work I am not sure if I made any mistakes:

I know that the terms ${n+1 \choose 2}$ occur exactly $n-times$ in the above expression so after regrouping terms I get \begin{align} s_n &=n{n+1 \choose 2}+(-1+0+1+2+\cdots+ (n-2))\\ & = -1+n{n+1 \choose 2} + (1+2+\cdots+ (n-2))\\ & = -1+n{n+1 \choose 2}+{n-1 \choose 2}\\ & = -1+{n^3+2n^2-3n+2 \above 1.5pt 2}\\ & ={n(n^2+2n-3) \above 1.5pt 2}\\ & ={n(n+3)(n-1) \above 1.5 pt 2} \end{align}

The sequence is $s_n =5,18,42,80,135,210,308\ldots$ and we can compare that to this sequence A212343. The motivation here are the sums of the following sets $\{2,3\}$, $\{5,6,7\}$, $\{9,10,11,12\}$ and so on.

Answer 1

The calculation is correct. Using the summation symbol and a slightly different calculation we obtain \begin{align*} s_n&=\sum_{j=-1}^{n-2}\left(j+\binom{n+1}{2}\right)\\ &=\sum_{j=1}^n\left(j-2+\binom{n+1}{2}\right)\tag{1}\\ &=\left(\sum_{j=1}^nj\right)-2n+n\binom{n+1}{2}\\ &=\frac{1}{2}n(n+1)-2n+n\frac{1}{2}(n+1)n\tag{2}\\ &=\frac{1}{2}n\left(n^2+2n-3\right) \end{align*} in accordance with your result.

Comment:

• In (1) we shift the index $j$ to start from $1$.

• In (2) we use the summation formula $\sum_{j=1}^nj=\frac{1}{2}n(n+1)$.