IPMAT Indore 2023 · Quantitative Ability · Question 10

Step 1 — write f\left( k \right) in closed form. It is the sum of the first k+1 positive integers:

f\left( k \right)=\frac{\left( k+1 \right)\left( k+2 \right)}{2}

Step 2 — split the reciprocal into partial fractions.

\frac{1}{f\left( k \right)}=\frac{2}{\left( k+1 \right)\left( k+2 \right)}=2\left( \frac{1}{k+1}-\frac{1}{k+2}\right)

Step 3 — telescope the sum.

g\left( n \right)=2\sum_{k=1}^{n}{\left( \frac{1}{k+1}-\frac{1}{k+2}\right)}=2\left( \frac{1}{2}-\frac{1}{n+2}\right)=1-\frac{2}{n+2}

Every interior term cancels against its neighbour; only the first and last survive.

Step 4 — apply the condition.

1-\frac{2}{n+2}>\frac{99}{100}\Rightarrow \frac{2}{n+2}<\frac{1}{100}\Rightarrow n+2>200

n>198 \Rightarrow n=199

Check: g\left( 199 \right)=1-2/201=199/201\approx 0.99005>0.99 ✔, while g\left( 198 \right)=1-2/200=0.99 exactly — which does not exceed 99/100.

Answer: 199

Shortcut. Recognise the telescoping shape immediately. f\left( k \right) is a triangular number, so \frac{1}{f\left( k \right)} is twice a difference of reciprocals and

g\left( n \right)=1-\frac{2}{n+2}

The decisive detail is the word “exceeds”. At n=198 you get exactly 99/100 — equal, not greater. The strict inequality pushes the answer to 199, and answering 198 is precisely the mistake the question is designed to catch.

General fact worth memorising: \sum_{k=1}^{n}{\frac{2}{k\left( k+1 \right)}}=2\left( 1-\frac{1}{n+1}\right) — all triangular-reciprocal sums reduce to this shape.