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1

### WB JEE 2009

For each n $$\in$$ N, 23n $$-$$ 1 is divisible by

here N is a set of natural numbers.

A
7
B
8
C
6
D
16

## Explanation

Let P(n) = 23n $$-$$ 1

Putting n = 1

P(1) = 23 . 1 $$-$$ 1 = 7 which is divisible by 7

Putting n = 2

P(2) = 23 . 2 $$-$$ 1 = 26 $$-$$ 1 = 63 which is divisible by 7 and so on

Let P(k) = 23k $$-$$ 1 is also divisible by 7

$$\therefore$$ 23k $$-$$ 1 = 7P $$\Rightarrow$$ 23k = 7P + 1

P(k + 1) = 23(k + 1) $$-$$ 1 = 23k . 23 $$-$$ 1

= (7P + 1)8 $$-$$ 1 = 7 . 8P + 8 $$-$$ 1 = 7(8P + 1)

$$\therefore$$ By process of M.I. P(k) is divisible by 7.

So, 23n $$-$$ 1 is divisible by 7.

2

### WB JEE 2009

Product of any r consecutive natural numbers is always divisible by

A
r!
B
(r + 4)!
C
(r + 1)!
D
(r + 2)!

## Explanation

Product of r consecutive natural numbers

= 1 . 2 . 3 . 4 . ...... . r = r! which is divisible by r! .

3

### WB JEE 2008

The coefficient of x$$-$$10 in $${\left( {{x^2} - {1 \over {{x^3}}}} \right)^{10}}$$ is

A
$$-$$ 252
B
$$-$$ 210
C
$$-$$ (5!)
D
$$-$$ 120

## Explanation

Let $${T_{r + 1}}$$ contains coefficient of x$$-$$10 in $${\left( {{x^2} - {1 \over {{x^3}}}} \right)^{10}}$$

$$\therefore$$ $${T_{r + 1}} = {}^{10}{C_r}{({x^2})^{10 - r}}{\left( { - {1 \over {{x^3}}}} \right)^r}$$

$$= {}^{10}{C_r}{( - 1)^r}{x^{20 - 5r}}$$

Comparing $${x^{ - 10}} = {x^{20 - 5r}}$$

$$\Rightarrow - 10 = 20 - 5r \Rightarrow r = 6$$

$$\therefore$$ Coefficient $$= {}^{10}{C_6}{( - 1)^6} = 210$$

$$\therefore$$ no option is given.

4

### WB JEE 2008

If $${}^{16}{C_r} = {}^{16}{C_{r + 1}}$$, then the value of $${}^r{P_{r - 3}}$$ is

A
31
B
120
C
210
D
840

## Explanation

$$\because$$ $${}^{16}{C_r} = {}^{16}{C_{r + 1}}$$

$$\Rightarrow {}^{16}{C_r} = {}^{16}{C_{15 - r}} \Rightarrow r = 15 - r$$

$$\Rightarrow r = 7.5$$ (not possible since r is whole number)

So no option is given.

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