IPMAT Indore 2025 · Quantitative Ability · Question 16

Step 1 — split the given sum into odds and evens.

\underset{\text{all}}{\underset{⏟}{\frac{1}{{1}^{2}}+\frac{1}{{2}^{2}}+\frac{1}{{3}^{2}}+\cdots}}=\underset{\text{odd}}{\underset{⏟}{\left( \frac{1}{{1}^{2}}+\frac{1}{{3}^{2}}+\cdots \right)}}+\underset{\text{even}}{\underset{⏟}{\left( \frac{1}{{2}^{2}}+\frac{1}{{4}^{2}}+\cdots \right)}}

Step 2 — the even part is a scaled copy of the whole thing. Every even number is 2k, so

\sum_{k\ge 1}{\frac{1}{{\left( 2k \right)}^{2}}}=\frac{1}{4}\sum_{k\ge 1}{\frac{1}{{k}^{2}}}=\frac{1}{4}\cdot \frac{{\pi}^{2}}{6}=\frac{{\pi}^{2}}{24}

This is the key move — the evens are just 1/4 of the total.

Step 3 — subtract.

\text{odd sum}=\frac{{\pi}^{2}}{6}-\frac{{\pi}^{2}}{24}=\frac{4{\pi}^{2}-{\pi}^{2}}{24}=\frac{3{\pi}^{2}}{24}=\frac{{\pi}^{2}}{8}

Answer: Option 4 — \frac{{\pi}^{2}}{8}