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 $18 s_{n} σ_{n-1}-s_{n}+2=0$

#Algebra

Easy Math Editor

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Comments

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

Pi Han Goh - 5 years, 6 months ago

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

Calvin Lin Staff - 5 years, 6 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.

Kay Xspre - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$