NIMCET 2010 — Mathematics PYQ

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Question

NIMCET 2010 — Mathematics PYQ

NIMCET | Mathematics | 2010

$\sum_{k=1}^{n} \frac{1}{k(k+1)}$ is equal to:

Choose the correct answer:

Explanation

Solving:

1. Use Partial Fractions:

The general term $T_k$ can be split into two parts:

T_k = \frac{1}{k(k+1)} = \frac{(k+1) - k}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}

2. Summing the Series (Telescoping Sum):

Let $S_n$ be the sum of the first $n$ terms:

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

Expanding the terms:

For $k=1$ : $\left( 1 - \frac{1}{2} \right)$
For $k=2$ : $\left( \frac{1}{2} - \frac{1}{3} \right)$
For $k=3$ : $\left( \frac{1}{3} - \frac{1}{4} \right)$
...
For $k=n$ : $\left( \frac{1}{n} - \frac{1}{n+1} \right)$

3. Cancellation:

Notice that the second part of each term cancels out the first part of the following term:

S_n = 1 - \cancel{\frac{1}{2}} + \cancel{\frac{1}{2}} - \cancel{\frac{1}{3}} + \cancel{\frac{1}{3}} - \dots + \cancel{\frac{1}{n}} - \frac{1}{n+1}

S_n = 1 - \frac{1}{n+1}

4. Simplify:

S_n = \frac{(n+1) - 1}{n+1} = \frac{n}{n+1}

Correct Option:

(c) $\frac{n}{n+1}$