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The sum of n−n- 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

Solution

✅ 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:

1k×(k+1)=1k−1k+1\frac{1}{k×(k+1)} = \frac{1}{k} - \frac{1}{k+1}

Sum = (11−12)+(12−13)+(13−14)+...+(1n−1n+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: 11−1n+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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