An algebra problem by swastik p (4)

Algebra Level 3

Let the sum ∑ k = 1 9 1 k ( k + 1 ) ( k + 2 ) \displaystyle \sum^{9}_{k=1}\dfrac{1}{k(k+1)(k+2)} written in its lowest terms be p q \dfrac{p}{q} . What is the value of q − p q-p ?

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The answer is 83.

S P by s p , 16, India

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1 solution

S P
Jul 29, 2018

∑ k = 1 9 1 k ( k + 1 ) ( k + 2 ) = ∑ k = 1 9 1 2 ( 1 k ( k + 1 ) − 1 ( k + 1 ) ( k + 2 ) ) = 1 2 [ ∑ k = 1 9 ( 1 k ( k + 1 ) − 1 ( k + 1 ) ( k + 2 ) ) ] = 1 2 ( 1 1 × 2 − 1 2 × 3 + 1 2 × 3 − ⋯ + 1 9 × 10 − 1 10 × 11 ) = 1 2 ( 1 2 − 1 110 ) = 1 2 × 54 110 = 27 110 \begin{aligned}\displaystyle \sum_{k=1}^{9}\dfrac{1}{k(k+1)(k+2)}&=\displaystyle \sum_{k=1}^{9}\dfrac{1}{2}\left(\dfrac{1}{k(k+1)}-\dfrac{1}{(k+1)(k+2)}\right)\\& =\dfrac{1}{2}\left[\displaystyle \sum_{k=1}^{9}\left(\dfrac{1}{k(k+1)}-\dfrac{1}{(k+1)(k+2)}\right)\right] \\& =\dfrac{1}{2}\left(\dfrac{1}{1\times 2}-\dfrac{1}{2\times 3}+\dfrac{1}{2\times 3}-\cdots+\dfrac{1}{9\times 10} -\dfrac{1}{10\times 11}\right) \\& =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{110} \right)\\& =\dfrac{1}{2}\times \dfrac{54}{110}\\&= \dfrac{27}{110}\end{aligned}

So, p q = 27 110 \dfrac{p}{q}=\dfrac{27}{110} and here p = 27 ; q = 110 p=27;q=110

Thus, q − p = 110 − 27 = 83 q-p=110-27=\boxed{83}

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