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Say there is sequnce of number $A(i) = i * (i+1)$, i.e. like 1*2, 2*3, 3*4 and so on. Is there any formula that allows to compute sum of the first N such numbers?

Gerry Myerson
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ivan.ukr
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  • Yes. Why don't you calculate the first few sums, look for a pattern, then try to prove it? – Gerry Myerson Feb 29 '16 at 23:26
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    If you rewrite the summation, you can get a formula in terms of the sum of the first $n$ natural numbers, and the sum of the squares of the first $n$ natural numbers. Recall $\sum\limits_{i=1}^ni=\frac{n(n+1)}{2},$ and

    $\sum\limits_{i=1}^ni^2=\frac{n(n+1)(2n+1)}{6}$

     – Peter Woolfitt Feb 29 '16 at 23:27

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Use that $$ k(k+1)(k+2)-(k-1)k(k+1)=k(k+1)·((k+2)-(k-1))=3·k(k+1) $$ to find a representation as telescoping sum.

This can be rewritten as $\binom{k+2}{3}-\binom{k+1}{3}=\binom{k+1}{2}$.

One can use the identity of Pascals triangle in general in an analogous way, as $$ \binom{k+m-1}m-\binom{k+m-2}m=\binom{k+m-2}{m-1}=\frac{k(k+1)···(k+m-2)}{(m-1)!} $$

Lutz Lehmann
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