JIPMAT 2022 (QA) - The sum of n- terms of sequence 1/1 x 2+1/2 x 3+1/3 x 4 . Is | PYQs + Solutions | AfterBoards
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JIPMAT 2022 (QA) PYQs

JIPMAT 2022

Algebra
>
Progression & Series

Easy

The sum of nn- terms of sequence 11×2+12×3+13×4\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4} \ldots \ldots. Is

Correct Option: 4
Looking at first few terms of sequence:
11×2+12×3+13×4+...+1n×(n+1)\frac{1}{1×2} + \frac{1}{2×3} + \frac{1}{3×4} + ... + \frac{1}{n×(n+1)}
Each term can be split using partial fractions:\newline1k×(k+1)=1k1k+1\frac{1}{k×(k+1)} = \frac{1}{k} - \frac{1}{k+1}
Sum = (1112)+(1213)+(1314)+...+(1n1n+1)(\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + ... + (\frac{1}{n} - \frac{1}{n+1})
Cancelling terms: 111n+1=nn+1\frac{1}{1} - \frac{1}{n+1} = \frac{n}{n+1}
Therefore, sum of n terms = nn+1\frac{n}{n+1}

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