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2 Answers

Best answer
7 votes
7 votes

The series can be written as:

$\displaystyle \begin{align}S &= \sum_{i=1}^{100} \frac{1}{i \cdot (i+1)}\\[2em] &= \sum_{i=1}^{100} \left ( \frac{1}{i} - \frac{1}{i+1}\right ) \end{align}$

We have written it in the form of a telescoping series.

Now, by expanding and cancelling terms we get:

$$\require{cancel} \begin{align}S &= \left ( \frac1 1 - \frac1 2 \right ) + \left ( \frac1 2 - \frac1 3 \right ) + \cdots + \left ( \frac1{99} - \frac1{100} \right ) + \left ( \frac1{100} - \frac1{101} \right )\\[3em] &= \frac1 1 + \cancelto{0}{\left ( \frac1 2 - \frac1 2\right )} + \cancelto{0}{\left ( \frac1 3 - \frac1 3\right )}+ \cdots + \cancelto{0}{\left ( \frac1{100} - \frac1{100}\right )} - \frac1{101}\\[3em]S &= \frac1 1 - \frac1{101}\\[1.5em] &= \frac{100}{101}\end{align}$$

Hence, option d) None of these is the correct answer.

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2 votes
2 votes

11.2+12.3+13.4....+1100.101

1/1*2 + 1/2*3 +........+ 1/100*101

= ( 1/1- 1/2) + (1/2-1/3) + ( 1/3-1/4)+ ......+(1/100-1/101)

= 1/1 - 1/101 = 100/101 

so ans is D .

11.2+12.3+13.4....+1100.101

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