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QUESTION: 1

Coefficient of x^{5} in the expansion of

Solution:

Coefficient of x^{5 }in (1 + x)^{10}

Coefficient of x^{25 }in (1 + x)^{30}

QUESTION: 2

The coefficients of 9^{th}, 10^{th} and 11^{th} terms in the expansion of (1 + x)^{n} are in A.P. then n =

Solution:

(n - 2r)^{2} = n + 2

QUESTION: 3

Coefficient of x^{4} in (1 + x – 2x^{2})^{6} is

Solution:

Factorise and expand

QUESTION: 4

Solution:

Consider the given function:

c1/c0 + 2c2/c1 + 3c3/c2 + 4c4/c3 +......+ncn/cn−1

=n/1 + [2n(n−1)]/2! . 1/n + [3n(n−1)(n−2)]/3! . 1/[n(n−1)/2!] +......+n.1/n

=n+(n−1)+(n−2)+......1

=1+2+3+......+n

= n(n+1)/2

QUESTION: 5

= (where ^{n}C_{r} = ^{25}C_{r})

Solution:

QUESTION: 6

It is suitable to use Binomial Distribution only for

Solution:

As the value of ‘n’ increases, it becomes difficult and tedious to calculate the value of ^{n}C_{x}.

QUESTION: 7

The number of terms in the expansion of [(a + 4b)^{3} + (a - 4b)^{3}]^{2} are

Solution:

[2(a^{3} + 3.a.16^{2})]^{2}

QUESTION: 8

If (1+x+x^{2})^{n} then a_{1} - 2a_{2} + 3a_{3} - ... - 2na_{2n} = ....

Solution:

QUESTION: 9

The sum of the binomial coefficients of the 3^{rd}, 4^{th} terms from the beginning and from the end of (a + x)^{n} is 440 then n =

Solution:

Applying the above conditions, we get

nC2 + nC3+ nCn−2 + nCn−3 = 440

2(nC2 + nC3)=440

nC2 + nC3=220

n(n−1)/2!+ [n(n−2)(n−1)]/3!=220

n(n−1)/2 (1+(n−2)/3=220

n(n−1)/2 (n+1)/3=220

n(n^{2}−1)/6 = 220

n^{3}−n=1320

n^{3}−n−1320=0

The only real root of the above equation is 11.

Hence n=11

QUESTION: 10

Let R = (5√5 + 11)^{2n+1}, f = R - [R], then Rf =

Solution:

Given, G = (5√5 + 11)^{2n + 1} , 0 < G < 1 and R = (5√5 + 11)^{2n + 1
}

QUESTION: 11

I + F when I is odd and 0 < F < 1, then (I + F) (I - F) =

Solution:

QUESTION: 12

The expansion [x + (x^{3} - 1)^{1/2}]^{5} + [x + (x^{3} - 1)^{1/2}]^{5 }is a polynomial of degree

Solution:

Expand using the formula

QUESTION: 13

If t_{0}, t_{1}, t_{2}, ............tn are the consecutive terms in the expansion (x + a)n then (t_{0} - t_{2} + t_{4} - t_{6} + ....)^{2} + (t_{1} - t_{3} + t_{5}....)^{2} =

Solution:

Expand (x + ai)^{n} and (x – ai)^{n} then multiply.

QUESTION: 14

Coefficient of x^{50} in (1 + x)^{1000} + 2x(1 + x)^{999} + 3x^{2} (1 +x)^{998} +....is

Solution:

Take (1 + x)^{1000} as common, after simplification it becomes (1 + x)^{1000} coefficient of x^{50} is ^{1002}C_{50}

QUESTION: 15

The coefficient of x^{9} in (x + 2) (x + 4) (x + 8).....(x + 1024) is

Solution:

2 + 4 + 8 + …..1024

QUESTION: 16

The coefficient of x^{n} in the polynomial (x +^{n} C_{0}) (x + 3.^{n}C_{1}) (x + 5.^{n}C_{2})...[x + (2n+ 1).^{n}C_{n}] is

Solution:

Coefficient of

QUESTION: 17

If x = (2 +√3)^{n}, n ∈ N and f = x - [x], then

Solution:

**Correct Answer :- d**

**Explanation : **x = (2 + (3)^{1/2})^{n}

For n = 1

x = 2 + (3)^{1/2}

f = x - [x]

=> [2 + (3)^{1/2}] - [2 + (3)^{1/2}]

=> (3)^{1/2} - 1

f^{2} = [(3)^{1/2} - 1]^{2}

=> 2(2 - (3)^{1/2})

1 - f = 1 - ((3)^{1/2} - 1)

1 - f = (2 - (3)^{1/2})

f^{2}/(1-f) = [2(2-(3)^{1/2})]/[(2-(3)^{1/2})]

For n = 2

f = (7 + 4(3)^{1/2}) - (7 + 4(3)^{1/2})

f = 7 + 4(3)^{1/2} - 13

f = 4(3)^1/2 - 6

f^{2} = [4(3)^{1/2} - 6]^{1/2}

=> 48 + 36 - 48(3)^{1/2}

=> 84 - 48(3)^{1/2}

1 - f = 1 - (48(3)^{1/2} - 6)

1 - f = 7 - 4(3)^{1/2}

f^{2}/(1-f) = [84 - 48(3)^{1/2}]/[7 - 4(3)^{1/2}]

= 12(Even)

QUESTION: 18

If 2^{2006} - 2006 divided by 7, the remainder is

Solution:

2^{2006} = 4(8)^{668} = 4 (7 + 1)^{688} leaves remainder 4

2006 = 7 x 268 + 4 leaves remainder

∴ 2^{2006} - 2006 leaves remainder

QUESTION: 19

In a Binomial Distribution, if p, q and n are probability of success, failure and number of trials respectively then variance is given by

Solution:

For a discrete probability function, the variance is given by

Where µ is the mean, substitute P(x)=^{n}C_{x} p^{x} q^{(n-x)} in the above equation and put µ = np to obtain

V = npq

QUESTION: 20

The coefficient of the term independent of x in the expansion of

Solution:

for this term to be independent of x. we must have

So, the required coefficient is

^{10}C_{4}(-1)^{4} = 210

QUESTION: 21

The coefficient of x^{12} in the expansion of (1+2x^{2} - x^{3})^{8}

Solution:

QUESTION: 22

Solution:

QUESTION: 23

The value of

Solution:

^{95}C_{4} + ^{95}C_{3} + ^{96}C_{3} + ^{97}C_{3} + ^{98}C_{3} + ^{98}C_{4} + ^{99}C_{3}

= ^{96}C_{4} + ^{96}C_{3} + ^{97}C_{3} + ^{98}C_{3} + ^{99}C_{3}

= ^{97}C_{4} + ^{97}C_{3} + ^{98}C_{3} + ^{99}C_{3}

= ^{98}C_{4} + ^{98}C_{3} + ^{99}C_{3} = ^{99}C_{4} + ^{99}C_{3} = ^{100}C_{4}

QUESTION: 24

Coefficient of x^{11} in the expansion of (1+x^{2})^{4} (1+x^{3})^{7} (1+x^{4})^{12} is

Solution:

QUESTION: 25

The number of irrational terms in the expansion of (2^{1/5} +3^{1/10})^{55} is

Solution:

(2^{1/5} +3^{1/10})^{55} is

Total term = 55 + 1 = 56

Here r = 0, 10, 20, 30, 40, 50

Number of rational terms = 6;

Number of irrational terms = 56 - 6 = 50

QUESTION: 26

In the expansion of , the coefficient of x^{-10} will be

Solution:

Given expansion is

∴ General term

Since, we have to find coefficient of x^{-10} ∴ -12 + 2r = -10 ⇒ r = 1

Now, then coefficient of x ^{-10} is 12 C_{1}(a)^{11} (b)^{1} = 12a^{11}b

QUESTION: 27

The middle term in the expansion of (1 – 2x + x^{2})^{n} is

Solution:

Here 2n is even integer, therefore,

term will be the middle term.

Now, (n + 1)^{th} term in

QUESTION: 28

If (1 - x + 2x^{2})^{10} equlas

Solution:

Replacing x by -x,

Subtracting and putting

QUESTION: 29

(103)^{86} - (86)^{103} - 86 is divisible by

Solution:

Given expression

QUESTION: 30

The number of dissimilar terms in the expansion of (a + b + c)^{2n+1} – (a + b – c)^{2n+1} is

Solution:

Given expansion

Number of dissimilar term = (2n + 1) + (2n -1) + ....(2n - 3) + .... + 5 + 3 + 1

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