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QUESTION: 1

The sum of coefficients of (1 + x - 3x^{2})^{2134} is

Solution:

For the sum of coefficient, put x = 1, to obtain the sum

which implies (1 + 1 - 3)^{2134} = 1

QUESTION: 2

The sum ^{r}C_{r} + ^{r+1}C_{r }+ ^{r+2}C_{r} + ....+^{n}C_{r} (n __>__ r) equals

Solution:

C(n, r) + c(n -1, r) + C(n - 2, r) + . . . + C(r, r)

= ^{r+1}C_{r+1} + ^{r+1}C_{r} + ^{r+2}C_{r} + . . . . + ^{n-1}C_{r} + ^{n}C_{r}

= ^{n+1}C_{r+1 }(applying same rule again and again)

(∴ ^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r})

QUESTION: 3

The expansion [x^{2} + (x^{6} - 1)^{1/2}]^{5} + [x^{2} - (x^{6} - 1)^{1/2}]^{5} is a polynomial of degree

Solution:

(x^{2} + (x^{6} - 1)^{1/2})^{5} + (x^{2} - (x^{6} - 1)^{1/2})^{5}

= 2(^{5}C_{0}(x^{2})^{5} + ^{5}C_{2} (x^{2})^{3} (x^{6} - 1) + ^{5}C_{4} (x^{6} - 1)^{2})

Here last term is of 14 degree.

QUESTION: 4

The term independent of x in

Solution:

The general term

The term independent of x, (or the constant term) corresponds to x^{18-3r} being x^{0} or 18 - 3r = 0 ⇒ r = 6 .

QUESTION: 5

The value of the greatest term in the expansion of

Solution:

Hence t_{1} < t_{2} < t_{3} < t_{4} < t_{5} <t_{6} <t_{7} < t_{8} < t_{9} < t_{10}

Hence, t_{8} is the greatest term and its value is

QUESTION: 6

9^{n+1} - 8n- 9 is divisible by

Solution:

(1+ 8)^{n+1} = ^{n+1}C_{0} + ^{n+1}C_{1}(8)^{1}+ ^{n+1}C_{2}(8)^{2} + ^{n+1}C_{3} (8)^{3} + .... + ^{n+1}C_{n+1}(8)^{n+1} (1)

⇒ 9^{n+1} = 1+(n+1)x8+^{n+1}C_{2}(8)^{3} + ....+^{n+1}C_{n-1}(8)^{n+1}

⇒ 9^{n+1} - 8n - 9 = (8)^{2} [^{n+1}C_{2 }+ ^{n+1}C_{3}(8)+^{n+1}C_{4}(8)^{2}+...+^{n+1}C_{n+1}(8)^{n-1}]

⇒ 9^{n+1} - 8n - 9 = 64 x an integer ⇒ 9^{n+1} - 8n - 9 is divisible by 64.

QUESTION: 7

The first integral term in the expansion of

Solution:

∴ T_{r+1} = ^{9}C_{r} (3^{1/2})^{9-r} (2^{1/3})^{r} = ^{9}C_{r} 3^{(9-r)/2}.2^{r/3}

For first integral term for r = 3;

T_{3+1} = ^{9}C_{3}3^{3}.2^{1} i.e., T_{3+1} = T_{4} (4^{th }term)

QUESTION: 8

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}

Total term = 55 + 1 = 56

T_{r+1} = ^{55}C_{r }2^{(55-r)/5 }3^{r/10}

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

Number of rational terms = 6;

Number of rational terms = 56 - 6 = 50

QUESTION: 9

The number of terms in the expansion of (2x + 3y- 4z)^{n} is

Solution:

We have, (2 + 3 - 4z)^{n} = {2 + (3 - 4)}^{n}

= ^{n}C_{0} (2x)^{n} (3y - 4z)° + ^{n}C_{1} (2x)^{n-1}(3y-4z)^{1} + ^{n}C_{2} (2x)^{n-2} (3y - 4z)^{2} +.. + ^{n}C_{n-1}(2x)^{1} (3y - 4z)^{n-1} + ^{n}C_{2}(3y-4z)n

Clearly, the first term in the above expansion gives one term, second term gives two terms, third term gives three terms and so on.

So, Total number of term = 1 +2+3+...+n+(n+1) =

QUESTION: 10

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

Solution:

Given expansion is

∴ General term = T_{r+1} = ^{12}C_{r} (a/x)^{12-r} (bx)^{r} = ^{12}C_{r} (a)^{12-r} b^{r} x^{-12+2r}

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: 11

If (1 + ax)^{n} = 1 + 8x + 24x^{2} + ….., then the values of a and n are equal to

Solution:

∴ n = 4,a = 2

QUESTION: 12

The product of middle terms in the expansion of is equal to

Solution:

**Correct Answer :- A**

**Explanation : **(x + 1/x)^{11}, it has 12 terms

Therefore, there are two middle terms (6th and 7th)

T_{6}T_{7} = [^{11}C_{5}(x)^{6}(1/x)^{5}] * [^{11}C_{6}(x)^{5}(1/x)^{6}]

^{11}C_{5} ^{11}C_{6}

QUESTION: 13

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 (1-x)^{2n} = ^{2n}C_{n}(1)^{2n-n}(-x)^{n} = ^{2n}C_{n}(-x)^{n}

QUESTION: 14

The sum of the binomial coefficients in the expansion of (x^{-3/4} + ax 5/4)^{n} lies between 200 and 400 and the term independent of x equals 448. The value of a is

Solution:

QUESTION: 15

^{23}C_{0} + ^{23}C_{2} + ^{23}C_{4} +...+^{23}C_{22} equals

Solution:

Given sum = sum of odd terms

QUESTION: 16

Solution:

(1- x+2x^{2})^{10} = a_{0}+a_{1}x+a_{2}x^{2} +..a_{19}x^{19} + a_{20}x^{20}

Replacing x by-x, (1+x+ 2x^{2})^{10} = a_{0 }- _{1}x+a_{2}x^{2}...a_{19}x^{19} +a_{20}x^{20})

Subtracting and putting x = 1, 2^{10} - 4^{10} = 2(a_{1} +a_{3} +...+a_{19})

⇒

QUESTION: 17

The greatest coefficient in the expansion of (1 + x)^{2n + 2} is

Solution:

Here 2n + 2 is even

Greatest coefficient

QUESTION: 18

(^{n}C_{0})^{2 }+ (^{n}C_{1})^{2 }+ (^{n}C_{2})^{2} +...+ (^{n}C_{n})^{2} equals

Solution:

(1+x)^{n} = ^{n}C_{0} + ^{n}C_{1}x + ^{n}C_{2}x^{2} - ..+ ^{n}C_{n}x^{n}.... (i)

(x+1)^{n} = ^{n}C_{0}x^{n} + ^{n}C_{1}x^{n-1} + ^{n}C_{2}x^{n-2} + ..... +^{n}C_{n} ......(ii)

Multiply (i) and (ii) and consider the coefficient x^{n} of both sides, we have

coefficient of

QUESTION: 19

The value of C_{1} + 3C_{3} + 5C_{5} + 7C_{7} + ...., where C_{0} , C_{3} , C_{5} , C_{7} ,..... are binomial coefficients is

Solution:

We know (1+x)^{n} = C_{0} +C_{1}x+C_{2}x^{2} +C_{3}x^{3} + ...+ C_{n}x^{n}n(1+ x)^{r-1} = C_{1}+ 2C_{2}x + 3C_{3}x^{2} +4C_{4}x^{3} +...+nC_{n}x^{n-1}

Putting x =1 and —1, then add

n2^{n-1} = C_{1} + 2C_{2} +3C_{3} +4C_{4} +..+nC_{n}

0= C_{1 }- 2C_{2} + 3C_{3} - 4C_{4} +... ⇒ n2^{n-2}

C_{1} + 3C_{3} +5C_{5} +...

QUESTION: 20

Fractional part of

Solution:

= 8(1+31)^{15 }= 8{^{15}C_{0 }+ ^{15}C_{1}31+..+^{15}C_{15}(31)^{15}}

2^{78} = 8 + an integer multiple of 31;

QUESTION: 21

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

Solution:

Given expression

QUESTION: 22

2^{3n} - bn- a is divisible by 49 then (a, b) is

Solution:

QUESTION: 23

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

QUESTION: 24

The numbers of terms in the expansion of

Solution:

Where ao = sum of all absolute terms = 1 +^{100}C_{2} .2 + ....

Similarly a_{1} , a_{2} , ...a_{100} and b_{1} , b_{2} ...b_{100} are coefficients obtained after simplification.

∴ Total number of terms = 1 + 100 + 100 = 201

QUESTION: 25

The coefficient of x^{50} in the expansion of (1+x)^{1000} + 2x(1+x)^{999} + 3x^{2}(1+x)^{998} + ....+1001x^{1000} is

Solution:

Subtract above equations,

⇒

Coefficient of x^{50} in S = coefficient of x^{50} in

QUESTION: 26

The coefficient of term independent of x in the expansion of

Solution:

Given expression

⇒

which is independent of x if

Hence required coefficient

QUESTION: 27

The value of where [x] represents integral part of ‘x’ is

Solution:

where I is integer and 0 __<__ f <1.

Now,

∴ I + f + F is integer.

QUESTION: 28

If in the expansion of (1 + x)^{m}(1 – x)^{n}, the coefficient of x and x^{2} are 3 and -6 respectively, then m is

Solution:

Given m - n = 3

⇒

⇒ (m - n)^{2} - m + n = -12 ⇒ m + n = 9 + 12 = 21 (2) using (1)

Solving (1) and (2), we get m = 12.

QUESTION: 29

is equal to

Solution:

QUESTION: 30

Coefficient of t^{24} in (1+t^{2})^{12}(1+t^{12})(1+t^{24}) is

Solution:

t^{24} in (1+t^{2})^{12}(1+t^{12})(1+t^{24}) is

Here, coefficient of t^{24} in

⇒ coefficient of t^{24} in

⇒ coefficient of t^{24} in

coefficient of

coefficient of

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