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### IIT-JEE 2010 Paper 1 Offline

Numerical
Let $${S_k}$$= 1, 2,....., 100, denote the sum of the infinite geometric series whose first term is $$\,{{k - 1} \over {k\,!}}$$ and the common ratio is $${1 \over k}$$. Then the value of $${{{{100}^2}} \over {100!}}\,\, + \,\,\sum\limits_{k = 1}^{100} {\left| {({k^2} - 3k + 1)\,\,{S_k}} \right|\,\,}$$ is

## Explanation

We have

$${S_k} = {{\left( {{{k - 1} \over {k!}}} \right)} \over {\left( {1 - {1 \over k}} \right)}} = {1 \over {(k - 1)!}}$$

Now, $$\sum\limits_{k = 2}^{100} {\left| {({k^2} - 3k + 1){1 \over {(k - 1)!}}} \right|}$$

$$= \sum\limits_{k = 2}^{100} {\left| {{{{{(k - 1)}^2} - k} \over {(k - 1)!}}} \right|}$$

$$= \sum {\left| {{{k - 1} \over {(k - 2)!}} - {k \over {(k - 1)!}}} \right|}$$

$$= \left| {{2 \over {1!}} - {3 \over {2!}}} \right| + \left| {{3 \over {2!}} - {4 \over {3!}}} \right| + ....$$

$$= {2 \over {1!}} - {1 \over {0!}} + {2 \over {1!}} - {3 \over {2!}} - {4 \over {3!}} + .... + {{99} \over {98!}} - {{100} \over {99!}}$$

$$= 3 - {{100} \over {99!}}$$

Thus, $${{{{100}^2}} \over {\left| \!{\underline {\, {100} \,}} \right. }} + \sum\limits_{k = 1}^{100} {\left| {({k^2} - 3k + 1){S_k}} \right| = 3}$$

2

### IIT-JEE 1990

Numerical
If $${\log _3}\,2\,,\,\,{\log _3}\,({2^x} - 5)\,,\,and\,\,{\log _3}\,\left( {{2^x} - {7 \over 2}} \right)$$ are in arithmetic progression, determine the value of x.

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