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QUESTION: 1

Let n be a positive integer. Then of the following, the greatest term is

Solution:

i)(1+1/4n)^{4n}

= (1+1/4)^{4}

= (5/4)4 = 625/256

ii)(1+1/3n)^{3n}

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

= (4/3)^{3} = 64/27

iii)(1+1/2n)^{2n}

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

= 6/4 = 3/2

iv)(1+1/n)^{n}

=(1+1)^{1} = 2

QUESTION: 2

The sum of the series (1^{2 }+ 1) 1! + (2^{2} + 1).2! + (3^{2} + 1).3! ...+(n^{2} + 1). n! is

Solution:

QUESTION: 3

The last two digits of the number 3^{400} are

Solution:

QUESTION: 4

The sum of the co-efficients of all the even powers of x in the expansion of (2x^{2} – 3x + 1)^{11} is

Solution:

QUESTION: 5

The coefficient of x^{r}(0 < r < n – 1) in the expression (x + 2)^{n-}^{1} + (x + 2)^{n-2} . (x + 1) + (x + 2)^{n-3}. (x + 1)^{2} + .............+ (x + 1)^{n-} is

Solution:

QUESTION: 6

The co-efficient of x^{401} in the expansion of (1 + x + x^{2} + ........+x^{9})^{-1}, (|x| < 1) is

Solution:

*Multiple options can be correct

QUESTION: 7

Let (1 + x^{2})^{2} (1 + x)^{n} = A_{0} + A_{1}x + A_{2}x^{2} + ........... If A_{0}, A_{1}, A_{2} are in A.P. then the value of n is

Solution:

*Multiple options can be correct

QUESTION: 8

The number 101^{100} – 1^{ }is divisible by

Solution:

*Multiple options can be correct

QUESTION: 9

If the 6^{th} term in the expansion of when x = 3 is numerically greatest then the possible integral value(s) of n can be

Solution:

*Multiple options can be correct

QUESTION: 10

If where I, n are integers and 0 < f < 1, then

Solution:

QUESTION: 11

Positive integers can be represented as

Solution:

*Multiple options can be correct

QUESTION: 12

In the expansion of

Solution:

*Multiple options can be correct

QUESTION: 13

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

Solution:

*Multiple options can be correct

QUESTION: 14

The value of ^{n}C_{0} . ^{n}C_{n} +^{ n}C_{1} .^{ n}C_{n – 1} + ... +^{ n}C_{n} . ^{n}C_{0} is

Solution:

*Multiple options can be correct

QUESTION: 15

The value of r for which ^{30}C_{r }. ^{20}C_{0 }+ ^{30}Cr – 1 . ^{20}C_{1 }+...+ ^{30}C_{0 }. ^{20}C_{ r }is maximum, is

Solution:

*Multiple options can be correct

QUESTION: 16

The value of r(0 < r < 30) for which ^{20}C_{r }. ^{10}C_{0 }+ ^{20}Cr – 1 . ^{10}C_{1 }+...+ ^{20}C_{0 }. ^{10}C_{r }is minimum, is

Solution:

*Multiple options can be correct

QUESTION: 17

^{20}C_{r }. ^{10}C_{0 }+ ^{20}Cr – 1 . ^{10}C_{1 }+...+ ^{20}C_{0 }. ^{10}C_{r }is minimum, is

Solution:

QUESTION: 18

Column-I Column-II

(A) If l be the number of terms in the expansion of (P) O + T = 3

(1 + 5x + 10x^{2} + 10x^{2} + 5x^{4} + x^{5})^{20} and if unit's

place and ten's place digits in 3^{l }are O and T, then

(B) If l be the number of terms in the expansion (Q) O + T = 7

of and if unit's place and (R) O + T = 9

ten's place digits in 7^{l }are O and T, then

(C) If l be the number of terms in the

expansion of (1 + x)^{101 }(1 + x^{2} _ x)^{100} (S) T _ O = 7

and if unit's place and ten's place

digits in 9^{l} are O and T, then (T) O _ T = 7

subjective type

Solution:

QUESTION: 19

Find the coefficients

(i) x^{7} in (ii) x ^{_7} in

(iii) Find the relation between a and b, so that these coefficients are equal.

Solution:

**Correct Answer :- ab = 1**

**Explanation : **rth term in the expansion of (a + b)ⁿ is given by ^{n}C_{r}(a)^{r}(b)^{n-r}

To find the coefficient of x⁷ in (ax² + 1/bx)¹¹

rth term is given by ^{11}C_{r}(ax²)^{r }(1/bx)^{11-r}

= ^{11}C_{r} a^{r} (1/b)^{11-r} x^{(r-22+2r)}

=^{11}C_{r} a^{r} (1/b)^{11-r} x^{3r-22}

To find the coefficient of x⁷, 3r - 11 should be 7

=> 3r - 11 = 7

=> 3r = 18

=> r = 6

Thus the coefficient of x⁷ is 11C6 a⁶b⁻⁵

To find the coefficient of x⁻⁷ in (ax - 1/bx²)¹¹

rth term is given by ^{11}C_{r} = (ax)^{r}(-1/bx²)^{11-r}

= ^{11}C_{r} a^{r} (-1/b)^{11-r} x^{(r-22+2r)}

=^{11}C_{r} a^{r} (1/b)^{11-r} x^{3r-22}

To find the coefficient of x⁻⁷, 3r - 22 should be -7

=> 3r - 22 = -7

=> 3r = 15

=> r = 5

Thus the coefficient of x⁷ is ^{11}C(a)⁵(-1/b)⁶ = ^{11}C_{6} a⁵b⁻⁶

Given that these coefficients are equal

=>^{11}C_{6} a⁶b⁻⁵ = ^{11}C_{6} a⁵b⁻⁶

⇒ a = 1/b

⇒ ab = 1

*Answer can only contain numeric values

QUESTION: 20

If the coefficients of (2r + 4)^{th} , (r – 2)^{th} terms in the expansion of (1 + x)^{18} are equal, find r.

Solution:

In the expansion of (1+x)^{18}

coefficient of T_{2r+4} = ^{18}C_{2r+3}

coefficient of T_{r−2} = ^{18}C_{r−3}

Given,

^{18}C_{2r+3} = ^{18}C_{r−3}

⇒ 2r−3 = r+3 or 2r+3+r−3 = 18

⇒ r = −6 or 3r = 18 ⇒ r = 6

but r cannot be negative

∴ r = 6

QUESTION: 21

If the coefficients of the r^{th}, (r + 1)^{th} and (r + 2)^{th} terms in the expansion of (1 + x)^{14} are in A.P., find r.

Solution:

QUESTION: 22

Find the term independent of x in the expansion of

(a) (b)

Solution:

QUESTION: 23

Given that (1 + x + x^{2})^{n }= a_{0} + a_{1}x + a_{2}x^{2} + ...... +a_{2n}x^{2n}, find the values of

(i) a_{0} + a_{1} + a_{2} + ..... + a_{2n} ;

(ii) a_{0} _ a_{1} + a_{2} _ a_{3} ....... +a_{2n} ;

(iii) a_{0}^{2} _ a_{1}^{2} + a_{2}^{2} _ a_{3}^{2} + ... + a_{2n}^{2}

Solution:

QUESTION: 24

In a Binomial Distribution, if ‘n’ is the number of trials and ‘p’ is the probability of success, then the mean value is given by

Solution:

QUESTION: 25

Prove that :

^{n _ 1}C_{r}^{ }+ ^{n _ 2}C_{r}^{ }+ ^{n _ 3}C_{r}^{ }+ ............ + ^{r}C_{r}^{ }= ^{n}C_{r + 1}^{.}

Solution:

### Inside level 2

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### Inside level 2

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### Practice Paper 2 - Binomial Theorem

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