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QUESTION: 1

If 'a' be the sum of the odd terms & 'b' the sum of the even terms in the expansion of (1+x)^{n}, then (1–x^{2})^{n} equals

Solution:

Sum of odd terms in the expansion of (1+x)^{n }will be 2^{(n-1) }....(i)

Now

(1−x^{2})^{n} = ^{n}C_{0} − ^{n}C_{1} x2 + ^{n}C_{2} x4 + ....(−1)^{n} ^{n}C_{n} x^{2n} .....(n)

(1+x^{2})^{n} = ^{n}C^{0} + _{n}C_{1} x^{2} + ^{n}C_{2} x^{4} +... nCn x^{2n} .....(m)

Subtracting n from m, we get

(1+x^{2})^{n} − (1−x^{2})^{n} = 2[^{n}C_{1} x^{2} + ^{n}C_{3} x^{6} + .....]

Now let x=1.

Hence we get

2^{n} − 0 = 2[^{n}C_{1} + ^{n}C_{3}^{ }+ .....]

Or ^{n}C_{1 }+ nC3x^{2} + ^{n}C_{5} + ... = 2^{(n-1)}

Hence a = b

Now (1−x2)^{n}

=(1+x)^{n} (1−x)^{n}

→[(a+b)(a−b)]

=a^{2} − b^{2}.

QUESTION: 2

Given that the term of the expansion (x^{1/3} – x^{-1/2})^{15} which does not contain x is 5m where m ∈ N, then m equals

Solution:

QUESTION: 3

In the binomial (2^{1/3} + 3^{-1/3})^{n}, if the ratio of the seventh term from the beginning of the expansion to the seventh term from its end is 1/6, then n equals

Solution:

QUESTION: 4

If the coefficients of x^{7} & x^{8} in the expansion of are equal, then the value of n is

Solution:

Since T_{r + 1 }= ^{n}C_{r }a^{n – r} x^{r} in expansion of (a + x)^{n},

Therefore,

QUESTION: 5

The expression is a polynomial in x of degree

Solution:

QUESTION: 6

Number of rational terms in the expansion of is

Solution:

QUESTION: 7

If n ∈ N & n is even, then equals

Solution:

1/1!(n−1)! + 1/3!(n−3) + 1/5!(n−5) +,

1/n![nC1 + nC3 + nC5 + …………….]

= 1/n![2^{(n-1)}]

QUESTION: 8

The sum of the binomial coefficients of is equal to 256. The constant term in the expansion is

Solution:

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

Sum of binomial coefficient = 2^{n} = 256

2^{n} = 2^{8} => n=8

8C4(2x)^{4}(1/x)^{4}

= [(8*7*6*5) * (2^{4}* x^{4}-4)]/1*2*3*4

= 1120

QUESTION: 9

The sum of the co-efficients in the expansion of (1 – 2x + 5x^{2})^{n} is 'a' and the sum of the co-efficients in the expansion of (1 + x)^{2n} is b. Then

Solution:

QUESTION: 10

The number of terms in the expansion of (x + y)^{n} is

Solution:

QUESTION: 11

The greatest terms of the expansion (2x + 5y)^{13} when x = 10, y = 2 is

Solution:

QUESTION: 12

The binomial expansion of , n ∈ N contains a term independent of x

Solution:

QUESTION: 13

Solution:

QUESTION: 14

Let where n ∈ N and p ∈ N and 0 < f < 1 then the value, f^{2} – f + pf – p is

Solution:

p→ integer f→ fraction

(5+2(6)^{½})^{n} = p+f {0<f<1}

f2 – f + pf – p

f(f-1) +p(f-1)

(f+p) (f-1)

(p+f) = (5+2(6)^{½})^{n }= nC0 5^{n} + nC1 5^{(n-1)}(2(6)^{½}) + nC2 5^{(n-2)}(2(6)^{½}) + nC3 5^{(n-3)}(2(6)^{½}) + ………………….nCn 5^{0} (2(6)^{½}) --------------------------(1)

0 < (5+2(6)^{½})^{n} which is approx

(5- 4.8989) = (0.11) < 1

0 <(0.11)^{n} <1

0 <(5-2(6)^{½})^{n} <1

0 <f’ <1

f’ = (5+2(6)^{½})^{n} = nC0 5^{n} - nC1 5^{(n-1)}(2(6)^{½}) + nC2 5^{(n-2)}(2(6)^{½}) - nC3 5^{(n-3)}(2(6)^{½}) + ………………….nCn 5^{0} (2(6)^{½}) --------------------------(2)

Adding (1) & (2)

p+f+f’ = 2[= (nC0 5^{n} + nC2 5^{(n-2})(2(6)^{½}) + nC4 5^{(n-4)}(2(6)^{½}) + …………………]

p+f=f’ = 2K K is integer

f+f’ = 2K-p = integer

f+f’ should be an integer

f^{2} - f + pf - p

= f(f-1) +p(f-1)

= (f+p) (f-1) = -(p+f) (1-f)

= -(p+f) (f’)

= - (5+2(6)^{½})^{n} (5-2(6)^{½})^{n}

= - {(5+2(6)^{½}) (5-2(6)^{½})}^{n}

= - (5)^{2} - (2(6)^{2})^{2}

= - [25 - 24]^{n} => - (1)^{n}

= - 1 (which is negative integer)

QUESTION: 15

If (1 + x + x^{2})^{25} = a_{0} + a_{1}x + a_{2}x^{2} + .........+a_{50}. x^{50} then a_{0} + a_{2} + a_{4} + ...........+ a_{50}is

Solution:

QUESTION: 16

The co-efficient of x^{4} in the expansion of (1 – x + 2x^{2})^{12} is

Solution:

QUESTION: 17

If then

Solution:

QUESTION: 18

Coefficient of a^{t} in the expansion of (a+p)^{m-1 }+ (a+p)^{m-2} (a+q) + (a+p)^{m-3} (a+q)^{2} + ....(a+q)^{m-1} where a > q and p > q is

Solution:

QUESTION: 19

Number of terms free from radical sign in the expansion of (1 + 3^{1/3} + 7^{1/7})^{10} is

Solution:

QUESTION: 20

In the expansion of (1 + x)^{n} (1 + y)^{n} (1 + z)^{n}, the sum of the co-efficients of the terms of degree 'r' is

Solution:

### Perfect Competition (Part - 1)

Doc | 10 Pages

### Inside level 1

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### Practice Paper 1 - Binomial Theorem

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- Test: Binomial Numbers (Competition Level)- 1
Test | 20 questions | 40 min

- Test: Binomial Numbers (Competition Level)- 2
Test | 25 questions | 50 min

- Test: Binomial Theorem (Competition Level)- 1
Test | 30 questions | 60 min

- Test: Binomial Theorem (Competition Level)- 3
Test | 30 questions | 60 min

- Test: Binomial Theorem (Competition Level)- 2
Test | 30 questions | 60 min