<s>[INST] The sum of $n$ terms in $1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots$ I am just confused, considering we can take $1 \cdot 2$ as the first term then we get the $n$th term as $n(n+1)$ so the sum of $n$ terms would be $\frac{n(n+1)(2n+1)}{6}$ + $\frac{n(n+1)}{2}$ but let's assume $0 \cdot 1$ as the first term then the $n$th term becomes $(n-1) \cdot n$ and so it's summation becomes $\frac{n(n+1) (2n+1)}{6} - \frac{n(n+1)}{2}$ but it's the same summation as above but the results are different what am I doing wrong?
 [/INST] This reduces the problem to the evaluation of a second derivative.
$$
   1\cdot 2 + 2\cdot 3+ 3\cdot 4+ \cdots + n\cdot (n+1)
   = \sum_{k=1}^{n} k(k+1) \\
   = \left.\frac{d^2}{dx^2}(1+x+x^2+x^3+\cdots+x^{n+1})\right|_{x=1} \\
   = \left.\frac{d^2}{dx^2}\frac{x^{n+2}-1}{x-1}\right|_{x=1}
$$
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