## Rajasthan Board RBSE Class 11 Maths Chapter 7 Binomial Theorem Miscellaneous Exercise

Question 1.

The number of terms in the expansion of

(A) 11

(B) 13

(C) 10

(D) 14

Solution :

Terms of R.H.S. of the expansion of (x + a)^{n} is finite for the positive values of x and number of terms is (n + 1).

So, value of n is 12 in the given expression.

Hence, number of total terms = 12 + 1 = 13.

Hence, option (B) is correct.

Question 2.

The 7th term in the expansion of

Solution :

7th term in the expansion of

Hence, option (C) is correct.

Question 3.

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

(A) 56a^{3} x^{5}

(B) -56a^{3 }x^{5}

(C) 70a^{4 }x^{4}

(D) -70a^{4 }x^{4}

Solution :

If n is even in (a + b)n, then its middle term

Question 4.

The constant term in the expansion

(A) 5th

(B) 4th

(C) 6th

(D) 7th

Solution :

(r + 1)th term in the expansion of

For constant term

x^{9} – 3r = x^{0}

⇒ 9 – 3r=0

⇒ r = 3

Hence, r + 1 = 3 + 1 = 4th term is the constant term

Hence, option (B) is correct, Ans.

Question 5.

The general term in the expansion of (x + a)^{n}

(A) ^{n}C_{r} x^{n-r }.a^{r}

(B) ^{n}C_{r} x^{r }.a^{r}

(C) ^{n}C_{n-r }x^{n-r}.a^{r}

(D) ^{n}C_{n-r }x^{r}.a^{n-r}

Solution :

General term in the expansion of (x + a)^{n}

⇒ T_{r}+1 = ^{n}C_{r} x^{n-r }a^{r}

Hence, option (A) is correct.

Question 6.

The value of term independent of x in the expansion

(A) 264

(B) -264

(C) 7920

(D) -7920

Solution :

(r + 1)th term in the expansion of

Question 7.

The coefficient of x^{-17} in the expansion of

(A) 1365

(B) -1365

(C) 3003

(D) -3003

Solution :

(x + 1)th term in the expansion of

Question 8.

If in the expansion of (1 + x)^{18}, coefficients of (2r + 4)th and (r – 2)th terms are equal then value of r is:

(A) 5

(B) 6

(C) 7

(D) 8

Solution :

In the expansion of (1 + x)^{18} coefficients of (2r + 4)th and (r – 2)th terms are ^{18}C_{2r + 3 }and ^{18}C_{r – 3} respectively

Given :

These are equal

Then, ^{18}C_{2r + 3}= ^{18}C_{r – 3}

⇒ 2r + 3 = r – 3

or 2r+ 3= 18 -(r – 3)

For using the statement if

^{n}C_{r} = ^{n}C_{p} or ^{n}C_{n-p}

⇒ 2r – r = -3 – 3

or 2r + 3 = 18 – r + 3

⇒ r = -6 or r = 6

r is a natural number, so r = – 6 is not possible, then x = 6.

Hence, option (B) is correct. .

Question 9.

If in the expansion of (a + b)^{n} and (a + b)^{n + 3} ratio of 2nd and 3rd, 3rd and 4th terms are equal, then value of n is :

(A) 5

(B) 6

(C) 3

(D) 4

Solution :

In the expansion of (a + b)^{n}

Hence, option (A) is correct.

Question 10.

If in the expansion of (1 + x)^{2n}, coefficient of 3rd and (r + 2)th term are equal, then :

(A) n = 2r

(B) n = 2r – 1

(C) n = 2r + 1

(D) n = r + 1

Solution :

In the expansion of (1 + x)^{2n}, 3rd term = 2^{n}C_{3r – 1 }and r + 2th term = 2^{n}C_{r + 1}

According to question,

2^{n}C_{3r – 1} = 2^{n}C_{r + 1}

Now, 3r – 1 = r + 1 or 3r – 1 = 2n – r – 1

⇒ 3r – r – I + 1 or 3r + r = 2n – 1 + 1

⇒ 2r = 2 or 4r = 2n

⇒ r =1 or 2r = n

Hence, option (A) is correct.

Question 11.

Find the value of term independent of x in the expansion of (2x – 1/x)^{10} :

Solution :

(r + 1)th term of the expansion of (2x – 1/x)^{10}

Hence, value of independent of x in the expansion of

Question 12.

Find the number of term in the expansion of (x + a)^{200}+ (x – a)^{200} after simplification.

Solution :

∵ (x + a)^{n} + (x – a)^{n}

= 2[^{n}C_{0} x^{n}a^{0} + ^{n}C_{0} x^{n-2} – a^{2} + ^{n}C_{4}x^{n-4 .}a^{4} +…]

where n is an even number then

[Number of terms in the expansion of (x + a)^{n}+ (x – a)^{n}]

Here, n = 200

Hence, number of terms in the expansion

Will be

Hence, number of terms is 101 in the expansion.

Question 13.

If the expansion of (1 + x)^{n}, C_{0} + C_{1} + C_{2} + C_{3} + … C_{n} are coefficients different terms then find the value C_{0} + C_{2} + C_{4}… .

Solution :

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

^{n}C_{2 }1^{n-2 }x^{2} + ^{n}C_{3 }1^{n-3}x^{3} +….

Putting x = 1

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

Putting x = – 1

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

Here ^{n}C_{0} + ^{n}C_{1} +^{n}C_{2} + ^{n}C_{3} + …. = 2^{n} … (i)

“C0 – “Ci + ”C2 – “C3 + … = 0 … (ii)

Adding equation (i) and (ii)

2[^{n}C_{0} + ^{n}C_{2} + ^{n}C_{4} + …] = 2^{n}

⇒ ^{n}C_{0} + ^{n}C_{2} + ^{n}C_{4} + … = 2^{n – 1
}or C_{0} + C_{2} + C_{4} + … = 2^{n – 1}

Question 14.

Find the value of

^{30}C_{1} + ^{30}C_{2} + ^{30}C_{3} +… + ^{30}C_{30} .

Solution :

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

Putting x = 1

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

or 2^{n} = ^{n}C_{0} + ^{n}C_{1} +^{n}C_{2} + ^{n}C_{3} + …. + ^{n}C_{n
}Here, putting n = 30.

2^{30} = 1+ ^{30}C_{1} +^{30}C_{2} + ^{30}C_{3} + …. + ^{30}C_{3}o

⇒ 1+ ^{30}C_{1} +^{30}C_{2} + ^{30}C_{3} + …. + ^{30}C_{3}o = 2^{30}

⇒ ^{30}C_{1} + ^{30}C_{2} + ^{30}C_{3} + … + ^{30}c_{30} = 2^{30} – 1

Question 15.

Find the middle term in the expansion of

Solution :

Question 16

In the product of expansion of (1 + 2x)^{6} (1 – x)^{7}, find the coefficient of x^{5}.

Solution :

Given expression

= (1 + 2x)^{6} (1 – x)^{7
}∵ (1 + 2x)^{6 }= ^{6}C_{0} + ^{6}C_{1} (2x)^{1} + ^{6}C_{2} (2x)^{2
}+ ^{6}C_{3} (2x)^{3} + ^{6}C_{4} (2x)^{4}+ ^{6}C_{5} (2x)^{5} + ^{6}C_{6} (2x)^{6
}= 1 + 6 × 2x + 15 × 4x^{2} + 20 × 8x^{3
}+ 15 × 16x^{4}+ 6 × 32 x 5+ 1 × 64x^{6
}or (1 + 2x)^{6} = 1 + 12x + 60x^{2} + 160x^{3
}+ 240x^{4}+ 192x^{5} + 64x^{6} … (i)

[∵ ^{6}C_{0} = 1, ^{6}C_{1} = 6, ^{6}C_{2} = 15, ^{6}C_{3} = 20, ^{6}C_{4} = 15, ^{6}C_{5} = 6]

and (1 – x)^{7} = ^{7}C_{0} – ^{7}C_{1}x + ^{7}C_{2}x^{2}– ^{7}C_{3}x^{3} + ^{7}C_{4}x^{4} – ^{7}C_{5}x^{6} +,..

= 1 – 7x + 21x^{2} – 35x^{3}+ 35x^{4}-21x^{5} +….. (ii)

[^{7}C_{0} = 1, ^{7}C_{1}= 7, ^{7}C_{2} = 21, ^{7}C_{3} = 35, ^{7}C_{4}= 35, ^{7}C_{5} = 21]

Now (1 + 2x)^{6} (1 – x)^{7
}= [1 + 12x + 60x^{2} + 160x^{3}+ 240x^{4} + 192x^{4}+ …]

= [1 – 7x + 21x^{2} – 35x^{3} + 35x^{4} – 21x^{5} + …]

Coefficient of x^{5} in the above expansion

[- 21 + 12 x 35 – 60 × 35 + 160 × 21+ 240 (-7) + 192 × 1]

= [- 21 + 420 – 2100 + 3360 – 1680 + 192]

= [-3801 + 3972]= 171.

Question 17.

If in the expansion of (1 + x)^{2n} coefficient of 2nd, 3nd and 4th terms are in A.P. then prove that 2n^{2} – 9n + 7 = 0.

Solution :

In the expansion of (1 + x)^{2n}, coefficient of 2nd, 3rd and 4th terms are

T_{2}= T_{1} + 1= ^{2n}C_{1
}T_{3} = T_{2} + 1 = ^{2n}C_{2
}T_{4 = }T_{3} +1 = ^{2n}C_{3
}According to questions,

⇒ 6 (2n – 1) = 6 + 2(2n – 1) (n – 1)

⇒ 12n – 6 = 6 + 2(2n^{2 }– 3n + 1)

⇒ 4 n^{2} – 6n + 2 – 12n + 6 + 6 = 0

⇒ 4n^{2 }– 18n +14 = 0

⇒ 2n^{2} – 9n + 7 = 0.

Hence Proved.

Question 18.

Solution :