If \(s_{n}\) denote the sum to \(n\) terms of the series \[1 \cdot 2+2 \cdot 3+ 3 \cdot 4+\ldots,\]and \(σ_{n-1}\) that to \(n-1\) terms of the series \[\dfrac{1}{1 \cdot 2 \cdot 3 \cdot 4}+\dfrac{1}{2 \cdot 3 \cdot 4 \cdot 5} +\ldots,\]show that \(18 s_{n} σ_{n-1}-s_{n}+2=0\)

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TopNewestFor the first question: See Sum of n, n², or n³.

For the second question: See the solutions to this problem. – Pi Han Goh · 1 year, 8 months ago

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– Calvin Lin Staff · 1 year, 8 months ago

I think the question that he's getting at is "Is there any way of showing that the final equation is true"?Log in to reply

First, find the partial sum of \(\sigma_n\). Note that

\[\frac{1}{n(n+1)(n+2)(n+3)} = \frac{1}{3}\:(\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)})\]

Now it is the telescopic series, to which its sum is

\[\sum_{m=1}^n \frac{1}{m(m+1)(m+2)(n+3)} = \frac{1}{3}\:(\frac{1}{6}-\frac{1}{(n+1)(n+2)(n+3)})\]

For sum up to \(n-1\) terms, the \(\sigma_{n-1}\) is then

\[\sigma_{n-1} = \frac{1}{3}\:(\frac{1}{6}-\frac{1}{n(n+1)(n+2)})\]

Now find \[s_n = \sum_{m=1}^n m(m+1) = \frac{n(n+1)}{2}(\frac{2n+1}{3}+1) = \frac{n(n+1)(n+2)}{3}\]

Therefore \(18\sigma_{n-1}s_n\) equals to \(s_n-2\) (multiplication omitted), and \(18\sigma_{n-1}s_n-s_n+2 = 0\). QED. – Kay Xspre · 1 year, 6 months ago

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