If 1/(2 3 4) + 1/(3 4 5) + 1/(4 5 6) + … + 1/(100 101 102) = k/101, then 34k is equal to _____.

JEE Main 2022 Mathematics question

If 1/(2 ) + 1/(3 ) + 1/(4 ) + … + 1/(100 ) = k/101, then 34k is equal to .

Answer: 286

Using 1/(n(n + 1)(n + 2)) = (1/2)[1/(n(n + 1)) − 1/((n + 1)(n + 2))], the sum telescopes: (1/2)[1/(2 ) − 1/(101 $×

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Question ID VK-035163