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I have an algorithm that generates terms as follows: $\begin{align*} \text{Term 1}: &2\times 3\times\left(\frac{1}{2}+\frac{1}{3}\right) + 4\\ \text{Term 2}: &2\times 3\times 4\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) + 3\times 5\\ \text{Term 3}: &2\times 3\times 4\times 5\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right) + 3\times4\times 6\\ \text{Term 4}: &2\times 3\times 4\times 5\times 6\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right) + 3\times4\times 5\times 7 \end{align*}$ and so on.

The pattern in the first half is obvious; the pattern in the second half is a little trickier. It "splits" the last multiplier into (multiplier-1)*(multiplier+1) so 4 changes into 3*5 and then the 5 on that one changes into 4*6 and so on for the next term.

I am trying to find a clever way to say "sum terms 1-N" given this pattern

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Term $k$ is $(k+2)!(H_{k+2}-1)+\frac12(k+1)!(k+3)\;,$ where $H_k$ is as usual the $k$-th harmonic number, $\sum_{i=1}^k\frac1k$, so the sum of the first $n$ terms is $\sum_{k=1}^n\Big((k+2)!(H_{k+2}-1)+\frac12(k+1)!(k+3)\Big)\;.$ You may of course replace $\frac12(k+1)!(k+3)$ by $\dfrac{(k+3)!}{2(k+2)}$ if you wish.

Added: Come to think of it, $n!H_n=\left[{n+1}\atop 2\right]$, a Stirling number of the first kind, so this can also be written $\begin{align*}\sum_{k=1}^n\left(\left[{k+3}\atop 2\right]-(k+2)!+\frac{(k+3)!}{2(k+2)}\right)&=\sum_{k=1}^n\left(\left[{k+3}\atop 2\right]-\frac{(k+1)(k+1)!}2\right)\;. \end{align*}$

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The last term is $\frac{(n+3)!}{2\times (n+2)},$ or $\frac{(n+1)!(n+3)}{2}$ where $n$ is the number of the term you are in.

So the $n$th term is: $\begin{align*} t_n &= (n+2)!\left(\frac{1}{2}+\cdots+\frac{1}{n+2}\right) + \frac{(n+3)!}{2\times(n+2)}\\ &= (n+2)!\left(\frac{1}{2}+\cdots+\frac{1}{n+2}\right) + \frac{(n+1)!(n+3)}{2}. \end{align*}$