 ### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2005

The sum of the series $$1 + {1 \over {4.2!}} + {1 \over {16.4!}} + {1 \over {64.6!}} + .......$$ ad inf. is
A
$${{e - 1} \over {\sqrt e }}\,$$
B
$${{e + 1} \over {\sqrt e }}$$
C
$${{e - 1} \over {2\sqrt e }}$$
D
$${{e + 1} \over {2\sqrt e }}$$

## Explanation

$${{{e^x} + {e^{ - x}}} \over 2}$$

$$= 1 + {{{x^2}} \over {2!}} + {{{x^4}} \over {4!}} + {{{x^6}} \over {6!}}.........$$

Putting $$x = {1 \over 2}$$ we get

$$1 + {1 \over {4.2!}} + {1 \over {16.4!}} + {1 \over {64.6!}} + .......$$

$$\infty = {{{e^{{1 \over 2}}} + {e^{{{ - 1} \over 2}}}} \over 2} = {{\sqrt e + {1 \over {\sqrt e }}} \over 2}$$

$$= {{e + 1} \over {2\sqrt e }}$$
2

### AIEEE 2005

If $$x = \sum\limits_{n = 0}^\infty {{a^n},\,\,y = \sum\limits_{n = 0}^\infty {{b^n},\,\,z = \sum\limits_{n = 0}^\infty {{c^n},} } } \,\,$$ where a, b, c are in A.P and $$\,\left| a \right| < 1,\,\left| b \right| < 1,\,\left| c \right| < 1$$ then x, y, z are in
A
G.P.
B
A.P.
C
Arithmetic-Geometric Progression
D
H.P.

## Explanation

$$x = \sum\limits_{n = 0}^\infty {{a^n}} = {1 \over {1 - a}}\,\,\,\,\,\,\,\,\,\,a = 1 - {1 \over x}$$

$$y = \sum\limits_{n = 0}^\infty {{b^n}} = {1 \over {1 - b}}\,\,\,\,\,\,\,\,\,\,b = 1 - {1 \over y}$$

$$z = \sum\limits_{n = 0}^\infty {{c^n}} = {1 \over {1 - c}}\,\,\,\,\,\,\,\,\,\,c = 1 - {1 \over z}$$

$$a,b,c$$ are in $$A.P.$$ OR $$2b = a + c$$

$$2\left( {1 - {1 \over y}} \right) = 1 - {1 \over x} + 1 - {1 \over y}$$

$${2 \over y} = {1 \over x} + {1 \over z} \Rightarrow x,y,z$$ are in $$H.P.$$
3

### AIEEE 2004

The sum of series $${1 \over {2\,!}} + {1 \over {4\,!}} + {1 \over {6\,!}} + ........$$ is
A
$${{\left( {{e^2} - 2} \right)} \over e}\,$$
B
$${{{{\left( {e - 1} \right)}^2}} \over {2e}}$$
C
$${{\left( {{e^2} - 1} \right)} \over {2e}}\,$$
D
$${{\left( {{e^2} - 1} \right)} \over 2}$$

## Explanation

We know that

$$e = 1 + {1 \over {1!}} + {1 \over {2!}} + {1 \over {3!}} + ..........$$

and

$${e^{ - 1}} = 1 - {1 \over {1!}} + {1 \over {2!}} - {1 \over {3!}} + .........$$

$$\therefore$$ $$e + {e^{ - 1}} = 2\left[ {1 + {1 \over {2!}} + {1 \over {4!}} + ......} \right]$$

$$\therefore$$ $${1 \over {2!}} + {1 \over {4!}} + {1 \over {6!}} + .......$$

$$= {{e + {e^{ - 1}}} \over 2} - 1$$

$$= {{{e^2} + 1 - 2e} \over {2e}}$$

$$= {{{{\left( {e - 1} \right)}^2}} \over {2e}}$$
4

### AIEEE 2004

The sum of the first n terms of the series $${1^2} + {2.2^2} + {3^2} + {2.4^2} + {5^2} + {2.6^2} + ....\,is\,{{n{{(n + 1)}^2}} \over 2}$$ when n is even. When n is odd the sum is
A
$${\left[ {{{n(n + 1)} \over 2}} \right]^2}$$
B
$${{{n^2}(n + 1)} \over 2}$$
C
$${{n{{(n + 1)}^2}} \over 4}$$
D
$$\,{{3n(n + 1)} \over 2}$$

## Explanation

If $$n$$ is odd, the required sum is

$${1^2} + {2.2^2} + {3^2} + {2.4^2} + ......$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2.{\left( {n - 1} \right)^2} + {n^2}$$

$$= {{\left( {n - 1} \right){{\left( {n - 1 + 1} \right)}^2}} \over 2} + {n^2}$$

[ As $$\left( {n - 1} \right)$$ is even

$$\therefore$$ using given formula for the sum of $$\left( {n - 1} \right)$$ terms.]

$$= \left( {{{n - 1} \over 2} + 1} \right){n^2} = {{{n^2}\left( {n + 1} \right)} \over 2}$$

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