If 1234+1345+1456++1100101102=k101, then 34k is equal to ________.

Question

If $\dfrac{1}{2	imes3	imes4}+\dfrac{1}{3	imes4	imes5}+\dfrac{1}{4	imes5	imes6}+\cdots+\dfrac{1}{100	imes101	imes102}=\dfrac{k}{101}$, then $34k$ is equal to \_\_\_\_\_\_\_\_.

Answer: 286

Accepted Answer

Using $\dfrac{1}{n(n+1)(n+2)} = \dfrac12\left(\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right)$, telescoping from $n=2$ to $100$:

$$S = \frac12\left(\frac16-\frac{1}{101\times102}\right) = \frac{143}{101\times17} = \frac{k}{101} \implies k=\frac{143}{17}.$$

$$34k = 34\times\frac{143}{17} = 2\times143 = 286.$$

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