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

Solution

Maths sir · 4 steps

Split each term into a difference of two-factor terms, and the whole sum collapses to first minus last.

1. Split the term. The two outer factors n and n+2 differ by 2. That lets you write each term as a difference of two neighbouring terms.

   1/n(n+1)(n+2) = 1/2[1/n(n+1) - 1/(n+1)(n+2)]

2. Telescope. Add this from n = 2 to n = 100. Every middle term cancels with the next one, leaving 1/2[1/(2 × 3) - 1/(101 × 102)] = 1/2[1/6 - 1/10302].
3. Simplify. 1/6 = 1717/10302, so the bracket is 1716/10302. Halving gives 858/10302 = 429/5151 = 429/(51 × 101).
4. So. Comparing with k/101 gives k = 429/51 = 143/17, so 34k = 2 × 143 = 286.

The common slip: 572. Forgetting the factor 1/2 in front of the split doubles the sum and gives 34k = 572.

The idea to keep. For 1/n(n+1)(n+2), subtract the two overlapping products and divide by the gap between the outer factors. Then only the first and last pieces survive.

Answer from the NTA final answer key. Solution by PrepWiser.