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1

### JEE Main 2021 (Online) 27th August Morning Shift

If 0 < x < 1, then $${3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + .....$$, is equal to :
A
$$x\left( {{{1 + x} \over {1 - x}}} \right) + {\log _e}(1 - x)$$
B
$$x\left( {{{1 - x} \over {1 + x}}} \right) + {\log _e}(1 - x)$$
C
$${{1 - x} \over {1 + x}} + {\log _e}(1 - x)$$
D
$${{1 + x} \over {1 - x}} + {\log _e}(1 - x)$$

## Explanation

Let $$t = {3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + ......\infty$$

$$= \left( {2 - {1 \over 2}} \right){x^2} + \left( {2 - {1 \over 3}} \right){x^3} + \left( {2 - {1 \over 4}} \right){x^4} + ......\infty$$

$$= 2({x^2} + {x^3} + {x^4} + .....\infty ) - \left( {{{{x^2}} \over 2} + {{{x^3}} \over 3} + {{{x^4}} \over 4} + .....\infty } \right)$$

$$= {{2{x^2}} \over {1 - x}} - (\ln (1 - x) - x)$$

$$\Rightarrow t = {{2{x^2}} \over {1 - x}} + x - \ln (1 - x)$$

$$\Rightarrow t = {{x(1 + x)} \over {1 - x}} - \ln (1 - x)$$
2

### JEE Main 2021 (Online) 26th August Morning Shift

If the sum of an infinite GP a, ar, ar2, ar3, ....... is 15 and the sum of the squares of its each term is 150, then the sum of ar2, ar4, ar6, ....... is :
A
$${5 \over 2}$$
B
$${1 \over 2}$$
C
$${25 \over 2}$$
D
$${9 \over 2}$$

## Explanation

Sum of infinite terms :

$${a \over {1 - r}} = 15$$ ..... (i)

Series formed by square of terms :

a2, a2r2, a2r4, a2r6 .......

Sum = $${{{a^2}} \over {1 - {r^2}}} = 150$$

$$\Rightarrow {a \over {1 - r}}.{a \over {1 + r}} = 150 \Rightarrow 15.{a \over {1 + r}} = 150$$

$$\Rightarrow {a \over {1 + r}} = 10$$ ...... (ii)

by (i) and (ii), a = 12; r = $${1 \over 5}$$

Now, series : ar2, ar4, ar6

Sum = $${{a{r^2}} \over {1 - {r^2}}} = {{12.\left( {{1 \over {25}}} \right)} \over {1 - {1 \over {25}}}} = {1 \over 2}$$
3

### JEE Main 2021 (Online) 26th August Morning Shift

The sum of the series

$${1 \over {x + 1}} + {2 \over {{x^2} + 1}} + {{{2^2}} \over {{x^4} + 1}} + ...... + {{{2^{100}}} \over {{x^{{2^{100}}}} + 1}}$$ when x = 2 is :
A
$$1 + {{{2^{101}}} \over {{4^{101}} - 1}}$$
B
$$1 + {{{2^{100}}} \over {{4^{101}} - 1}}$$
C
$$1 - {{{2^{100}}} \over {{4^{100}} - 1}}$$
D
$$1 - {{{2^{101}}} \over {{2^{400}} - 1}}$$

## Explanation

$$S = {1 \over {x + 1}} + {2 \over {{x^2} + 1}} + {{{2^2}} \over {{x^4} + 1}} + ...... + {{{2^{100}}} \over {{x^{{2^{100}}}} + 1}}$$

$$S + {1 \over {1 - x}} = {1 \over {1 - x}} + {1 \over {x + 1}} + ...... = {2 \over {1 - {x^2}}} + {2 \over {1 + {x^2}}} + ....$$

$$S + {1 \over {1 - x}} = {{{2^{101}}} \over {1 - {x^{400}}}}$$

$$S = 1 - {{{2^{101}}} \over {{2^{400}} - 1}}$$
4

### JEE Main 2021 (Online) 25th July Morning Shift

Let Sn be the sum of the first n terms of an arithmetic progression. If S3n = 3S2n, then the value of $${{{S_{4n}}} \over {{S_{2n}}}}$$ is :
A
6
B
4
C
2
D
8

## Explanation

Let a be first term and d be common diff. of this A.P.

Given, S3n = 3S2n

$$\Rightarrow {{3n} \over 2}[2a + (3n - 1)d] = 3{{2n} \over 2}[2a + (2n - 1)d]$$

$$\Rightarrow 2a + (3n - 1)d = 4a + (4n - 2)d$$

$$\Rightarrow 2a + (n - 1)d = 0$$

Now, $${{{S_{4n}}} \over {{S_{2n}}}} = {{{{4n} \over 2}[2a + (4n - 1)d]} \over {{{2n} \over 2}[2a + (2n - 1)d]}} = {{2\left[ {\underbrace {2a + (n - 1)d}_{ = 0} + 3nd} \right]} \over {\left[ {\underbrace {2a + (n - 1)d}_{ = 0} + nd} \right]}}$$

$$= {{6nd} \over {nd}} = 6$$

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