# How to solve this amusing problem?

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 $$18 s_{n} σ_{n-1}-s_{n}+2=0$$

Note by Rohit Udaiwal
2 years, 5 months ago

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For the first question: See Sum of n, n², or n³.
For the second question: See the solutions to this problem.

- 2 years, 5 months ago

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

Staff - 2 years, 5 months ago

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.

- 2 years, 4 months ago