I have a progression that goes like this:
$$1\times 2 + 2 \times 3 + .. + (n-1) \times n$$ Is there a way I can simplify it?
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$2\binom n3{}{}{}{}$ – abiessu May 26 '15 at 23:31
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Sum of the first $n$ triangular numbers - induction – Martin Sleziak May 27 '15 at 06:17
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Hint. You have $$ (n^3-n)-\left((n-1)^3-(n-1)\right)=3\times(n-1)\times n $$ and the sum is telescoping.
Olivier Oloa
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Can you explain how did you get from my progression to the left expression? – Max Yankov May 26 '15 at 23:43
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Yes, I recalled that $(n-1)^3=n^3-3n^2+3n-1$ giving $n^3-(n-1)^3=3\times (n-1)n+1$, then I obtained the given identity. Thanks. – Olivier Oloa May 26 '15 at 23:48
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I'm sorry, but your explanation still didn't show anything that looked like my series. – Max Yankov May 27 '15 at 09:13
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You are considering $$S=\sum_{i=1}^{n-1}i\,(i+1)=\sum_{i=1}^{n-1}i^2+\sum_{i=1}^{n-1}i$$ which are quite standard summations. For the first one $$\sum_{i=1}^{n-1}i^2=\frac{1}{6} (n-1) n (2 n-1)$$ and for the second $$\sum_{i=1}^{n-1}i=\frac{1}{2} (n-1) n$$ Add them together and simplify (noticing some nice factorizations).
Claude Leibovici
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Yes. \begin{align*} 1\cdot 2 + 2\cdot 3 + \dotsb + (n-1)\cdot n &= 1\cdot(1+1) + 2\cdot(2+1) + \dotsb + (n-1)\cdot((n-1)+1) \\ &= 1^2 + 1 + 2^2 + 2 + \dotsb + (n-1)^2 + (n-1) \\ &= (1^2+2^2+\dotsb+(n-1)^2) + (1+2+\dotsb+(n-1)) \end{align*} Both of those sums have well-known closed forms; see here, for example.
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You have $T_r=r(r+1)$ which you can write as $$T_r=\frac{r(r+1)\{(r+2)-(r-1)\}}{(r+2)-(r-1)}$$ $$T_r=\frac{1}{3}\{r(r+1)(r+2)-(r-1)r(r+1)\}$$ And there you have your telescoping series. On adding up all terms you will find $$S_{n-1}=\frac{1}{3}(n-1)n(n+1)$$
Arpan
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