Let the sum ∑_ n=1 ^9 (1)/(n(n+1)(n+2)) written in its lowest terms be (p)/(q) . Find the value of q-p .

Correct answer: 83 aligned & ∑_ n=1 ^9 (1)/(2) (n+2-n)/(n(n+1)(n+2)) & =(1)/(2) ∑_ n=1 ^9((1)/(n(n+1))-(1)/((n+1)(n+2))) & =(1)/(2)(((1)/(1.2)-(1)/(2.3))+((1)/(2.3)-(1)/(3.4))+ +((1)/(9 × 10)-(1)/(10 × 11))) & =(1)/(2)((1)/(2)-(1)/(110)) & =(1)/(2)((55-1)/(110)) (27)/(110) q-p=110-27=83 aligned

Olympiad workbook

Exam: Olympiad workbook
Subject: IOQM
Chapter: Sequences and Series

Olympiad workbook · IOQM · Sequences and Series

Let the sum ∑_ n=1 ^9 (1)/(n(n+1)(n+2)) written in its lowest terms...

Let the sum

\sum_{n=1}^9 \frac{1}{n(n+1)(n+2)}

written in its lowest terms be

\frac{p}{q}

. Find the value of

q-p

Correct answer

Step-by-step solution

\begin{aligned} & \sum_{n=1}^9 \frac{1}{2} \frac{n+2-n}{n(n+1)(n+2)} \\ & =\frac{1}{2} \sum_{n=1}^9\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right) \\ & =\frac{1}{2}\left(\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+\ldots \ldots \ldots+\left(\frac{1}{9 \times 10}-\frac{1}{10 \times 11}\right)\right) \\ & =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{110}\right) \\ & =\frac{1}{2}\left(\frac{55-1}{110}\right) \Rightarrow \frac{27}{110} \Rightarrow q-p=110-27=83\end{aligned}

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