Inter 1st Year Maths Binomial Theorem Solutions Exercise 7b

Practicing the AP Board Solutions Class 11 Maths and Chapter 7 Inter 1st Year Maths Binomial Theorem Solutions Exercise 7b Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Binomial Theorem Solutions Exercise 7b

Binomial Theorem Exercise 7b Solutions

Binomial Theorem Class 11 Exercise 7b Solutions – Binomial Theorem 7b Exercise Solutions

I.

Question 1.
If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer.
[Hint: Write aⁿ = (a – b + b)ⁿ and expand]
Solution:
To prove that (a – b) is a factor of (aⁿ – bⁿ), it has to be proved that aⁿ – bⁿ = k(a – b) where k is some natural number.
a can be written as a = a – b + b

This shows that (a – b) is a factor of (aⁿ – bⁿ), where n is a positive integer.

Question 2.
Evaluate (√3 + √2)⁶ – (√3 – √2)⁶
Solution:
Using the binomial theorem, the expressions (a + b)⁶ and (a – b)⁶ can be expanded

Question 3.
Find the value of \(\left(a^2+\sqrt{a^2-1}\right)^4+\left(a^2-\sqrt{a^2-1}\right)^4\).
Solution:
Firstly the expression (x + y)⁴ + (x – y)⁴ is simplified by using binomial theorem

II.

Question 1.
Find an approximation of (0.99)⁵ using the first three terms of its expansion.
Solution:
0.99 can be written as 0.99 = 1 – 0.01
Now, by applying the binomial theorem, we get
(0.99)⁵ = (1 – 0.01)⁵
= ⁵C₀ (1)⁵ – ⁵C₁ (1)⁴ (0.01) + ⁵C₂ (1)³ (0.01)²
= 1 – 5(0.01) + 10(0.01)²
= 1 – 0.05 + 0.001
= 0.951

Question 2.
Expand using Binomial Theorem \(\left(1+\frac{x}{2}-\frac{2}{x}\right)^4\), x ≠ 0.
Solution:
Using the binomial theorem, the given expression can be expanded as

Again, by using the binomial theorem to expand the above terms, we get

From equations 1, 2, and 3, we get

Question 3.
Find the expansion of (3x² – 2ax + 3a²)³ using the binomial theorem.
Solution:
We know that (a + b)³ = a³ + 3a²b + 3ab² + b³
Putting a = 3x², b = -a(2x – 3a), we get

Now multiplying and dividing by n, we get

From the above equations, L.H.S = R.H.S
Hence the proof.

Question 4.
Prove that C₀ + 3.C₁ + 5.C₂ + …… + (2n + 1).C_n = (2n + 2) . 2^n-1
Solution:
Let S = C₀ + 3.C₁ + 5.C₂ + ….. + (2n+1) . C_n ……….(1)
By writing the terms in (1) in the reverse order, we get

Question 5.
Prove that, for any real numbers a, d,
a.C₀ + (a + d).C₁ + (a + 2d).C₂ +….+ (a + nd).C_n = (2a + nd).2^n-1
Solution:

Question 6.
If n is a positive integer, then prove that \(C_0+\frac{C_1}{2}+\frac{C_2}{3}+\ldots .+\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}\)
Solution:
Write S = \(C_0+\frac{C_1}{2}+\frac{C_2}{3}+\ldots .+\frac{C_n}{n+1}\), then

Question 7.
If n is a positive integer and x is any non-zero real number, then prove that

Solution:

Question 8.
Prove that 2.C₀ + 5.C₁ + 8.C₂ +….+ (3n + 2)C_n = (3n + 4)2^n-1
Solution:

Question 9.
Prove that C₀ + 2.C₁ + 4.C₂ + 8.C₃ +….+ 2ⁿ . C_n = 3ⁿ
Solution:

Question 10.
If (1 + x + x²)ⁿ = a₀ + a₁ x + a₂ x² + …. + a_2n x²ⁿ, then prove that
(i) a₀ + a₁ + a₂ + ……… + a_2n = 3ⁿ
(ii) a₀ + a₂ + a₄ + ……. + a_2n = \(\frac{3^n+1}{2}\)
(iii) a₁ + a₃ + a₅ + …….. + a_2n-1 = \(\frac{3^n-1}{2}\)
(iv) a₀ + a₃ + a₆ + a₉ + ….. = 3^n-1
Solution:

Question 11.
If P and Q are the sum of odd terms and the sum of even terms, respectively, in the expansion of (x + a)ⁿ, then prove that
(i) P² – Q² = (x² – a²)ⁿ
(ii) 4PQ = (x + a)²ⁿ – (x – a)²ⁿ
Solution:

Question 12.
If (1 + 3x – 2x²)¹⁰ = a₀ + a₁ x + a₂ x² + …. + a₂₀ x²⁰, then prove that
(i) a₀ + a₁ + a₂ + …… + a₂₀ = 2¹⁰
(ii) a₀ – a₁ + a₂ – a₃ + …. + a₂₀ = 4¹⁰
Solution:
Given (1 + 3x – 2x²)¹⁰ = a₀ + a₁ x + a₂ x² + …. + a₂₀ x²⁰ ………..(1)
(i) Put x = 1 in equation (1)
(1 + 3 – 2)¹⁰ = a₀ + a₁ + a₂ + …… + a²⁰
∴ a₀ + a₁ + a₂ + ……….. + a₂₀ = 2¹⁰
(ii) Put x = -1 in equation (1)
(1 – 3 – 2)¹⁰ = a₀ – a₁ + a₂ + …. + a₂₀
∴ a₀ – a₁ + a₂ – a₃ + …….. + a₂₀ = (-4)¹⁰ = 4¹⁰

Question 13.
Show that the middle term in the expansion of (1 + x)²ⁿ is \(\frac{1.3 .5 \ldots .(2 n-1)}{n!}\) 2n . xⁿ, where n is a positive integer.
Solution:
The expansion of (1 + x)²ⁿ contains (2n + 1) terms.

Question 14.
The second, third, and fourth terms in the binomial expansion (x + a)ⁿ are 240, 720, and 1080, respectively. Find the x, a, and n.
Solution:



Hence x = 2, a = 3, n = 5