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SSC GD 2024 PYQ — Question 54

SSC GD Const (English) - 22 Feb 2024 - Shift 4 • Quantitative Aptitude • Q54

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What is the value of 21×3+23×5+25×7........299×101\frac{2}{1 × 3} + \frac{2}{3 × 5} + \frac{2}{5 × 7} ........ \frac{2}{99 × 101} ?

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Correct answer: C

Explanation

The correct answer is option 3  i.e. 100/101.

⇒ S = 21×3+23×5+25×7+⋯+299×101\frac{2}{1 \times 3} + \frac{2}{3 \times 5} + \frac{2}{5 \times 7} + \cdots + \frac{2}{99 \times 101}

⇒2(2n−1)(2n+1)=24n2−1\frac{2}{(2n-1)(2n+1)} = \frac{2}{4n^2 - 1}

⇒ 2(2n−1)(2n+1)=A2n−1+B2n+1\frac{2}{(2n-1)(2n+1)} = \frac{A}{2n-1} + \frac{B}{2n+1}

⇒ 2 = A(2n+1) + B(2n-1)

⇒ 2 = A(2n) + A + B(2n) - B

⇒ 2 = (2A + 2B)n + (A - B)

⇒ 2A + 2B = 0 , A - B = 2

⇒−B−B=2⇒−2B=2⇒B=−1-B - B = 2 \quad \Rightarrow \quad -2B = 2 \quad \Rightarrow \quad B = -1

⇒ 2(2n−1)(2n+1)=12n−1−12n+1\frac{2}{(2n-1)(2n+1)} = \frac{1}{2n-1} - \frac{1}{2n+1}

⇒S = ∑n=150(12n−1−12n+1)\sum_{n=1}^{50} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right)

⇒ S = (11−13)+(13−15)+(15−17)+⋯+(199−1101)\left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{7} \right) + \cdots + \left( \frac{1}{99} - \frac{1}{101} \right)

⇒ S = 1−11011 - \frac{1}{101}

⇒ S = 101101−1101=100101\frac{101}{101} - \frac{1}{101} = \frac{100}{101}

The value of the series is 100101\frac{100}{101}