Inter 1st Year Maths Binomial Theorem Solutions Exercise 7a

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

Intermediate 1st Year Maths Binomial Theorem Solutions Exercise 7a

Binomial Theorem Exercise 7a Solutions

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

I. Expand each of the expressions in Exercises 1 to 5.

Question 1.
(1 – 2x)⁵
Solution:
From binomial theorem expansion,
We can write as (1 – 2x)⁵

Question 2.
\(\left(\frac{2}{x}-\frac{x}{2}\right)^5\)
Solution:
From the binomial theorem, the given equation can be expanded as

Question 3.
(2x – 3)⁶
Solution:
From the binomial theorem, the given equation can be expanded as

Question 4.
\(\left(\frac{x}{3}+\frac{1}{x}\right)^5\)
Solution:
From the binomial theorem, the given equation can be expanded as

Question 5.
\(\left(x+\frac{1}{x}\right)^6\)
Solution:
From the binomial theorem, the given equation can be expanded as

Question 6.
(4x + 5y)⁷
Solution:

Question 7.
\(\left(\frac{2}{3} x+\frac{7}{4} y\right)^5\)
S0lution:

Question 8.
\(\left(\frac{2 p}{5}-\frac{3 q}{7}\right)^6\)
Solution:

Question 9.
Write down and simplify.
(i) 6th term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\)
Solution:
The general term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\) is

(ii) 7th term in (3x – 4y)¹⁰
Solution:
General term in (3x – 4y)¹⁰ is

(iii) 10th term in \(\left(\frac{3 p}{4}-5 q\right)^{14}\)
Solution:
General term in \(\left(\frac{3 p}{4}-5 q\right)^{14}\) is

(iv) rth term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) (1 ≤ r ≤ 8).
Solution:
The general term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) is

(v) Middle term in \(\left(3-\frac{x^3}{6}\right)^7\)
Solution:

(vi) Middle term in \(\left(\frac{x}{3}+9 y\right)^{10}\)
Solution:
Given \(\left(\frac{x}{3}+9 y\right)^{10}\)
Total terms = 10 + 1 = 11
Middle term

(vii) 4th term in (x – 2y)¹²
Solution:

(viii) 13th term in \(\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}\), x ≠ 0
Solution:

Question 10.
Find the number of terms in the expansion of
(i) \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\)
(ii) (3p + 4q)¹⁴
Solution:
(i) Number of terms in (x + a)ⁿ is (n + 1), where n is a positive integer.
Hence number of terms in \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\) are 9 + 1 = 10
(ii) Number of terms in (3p + 4q)¹⁴ are 14 + 1 = 15.

II. Using the binomial theorem, evaluate each of the following:

Question 1.
(96)³
Solution:
96 can be expressed as the sum or difference of two numbers, and then the binomial theorem can be applied.
The given question can be written as 96 = 100 – 4
(96)³ = (100 – 4)³
= ³C₀ (100)³ – ³C₁ (100)² (4) – ³C₂ (100) (4)³ – ³C₃ (4)²
= (100)³ – 3(100)² (4) + 3(100) (4)² – (4)³
= 1000000 – 120000 + 4800 – 64
= 884736

Question 2.
(102)⁵
Solution:
Given (102)⁵
102 can be expressed as the sum or difference of two numbers, and then the binomial theorem can be applied.
The given question can be written as 102 = 100 + 2
(102)⁵ = (100 + 2)⁵
= ⁵C₀ (100)⁵ + ⁵C₁ (100)⁴ (2) + ⁵C₂ (100)³ (2)² + ⁵C₃ (100)² (2)³ + ⁵C₄ (100) (2)⁴ + ⁵C₅ (2)⁵
= (100)⁵ + 5(100)⁴ (2) + 10(100)³ (2)² + 5(100)² (2)³ + 5(100) (2)⁴ + (2)⁵
= 1000000000 + 1000000000 + 40000000 + 80000 + 8000 + 32
= 11040808032

Question 3.
(101)⁴
Solution:
Given (101)⁴
101 can be expressed as the sum of the difference of two numbers; then, the binomial theorem can be applied.
The given question can be written as 101 = 100 + 1
(101)⁴ = (100 + 1)⁴
= ⁴C₀ (100)⁴ + ⁴C₁ (100)³ (1) + ⁴C₂ (100)² (1)² + ⁴C₃ (100) (1)³ + ⁴C₄ (1)⁴
= (100)⁴ + 4(100)³ + 6(100)² + 4(100) + (1)⁴
= 100000000 + 4000000 + 60000 + 400 + 1
= 104060401

Question 4.
(99)⁵
Solution:
Given (99)⁵
99 can be written as the sum or difference of two numbers, then the binomial theorem can be applied.
The given question can be written as 99 = 100 – 1
(99)⁵ = (100 – 1)⁵
= ⁵C₀ (100)⁵ – ⁵C₁ (100)⁴ (1) + ⁵C₂ (100)³ (1)² – ⁵C₃ (100)² (1)³ + ⁵C₄ (100) (1)⁴ – ⁵C₅ (1)⁵
= (100)⁵ – 5(100)⁴ + 10(100)³ – 10(100)² + 5(100) – 1
= 100000000 – 5000000000 + 10000000 + 500 – 1
= 9509900499

Question 5.
Using the Binomial Theorem, indicate which number is larger: 1.1)¹⁰⁰⁰⁰ or 1000.
Solution:
By splitting the given 1.1 and then applying the binomial theorem,
The first few terms of (1.1)¹⁰⁰⁰⁰ can be obtained as
(1.1)¹⁰⁰⁰⁰ = (1 + 0.1)¹⁰⁰⁰⁰
= 1 + 10000 × 1.1 + other positive terms
= 1 + 11000 + other positive terms
>1000
(1.1)¹⁰⁰⁰⁰ > 1000

Question 6.
Find (a + b)⁴ – (a – b)⁴. Hence, evaluate (√3 + √2)⁴ – (√3 – √2)⁴.
Solution:
Using the binomial theorem,
The expressions (a + b)⁴ and (a – b)⁴ should be expanded.

Question 7.
Find (x + 1)⁶ + (x – 1)⁶. Hence or otherwise evaluate (√2 + 1)⁶ + (√2 – 1)⁶.
Solution:
Using binomial theorem, the expressions (x + 1)⁶ and (x – 1)⁶ can be expressed as

Question 8.
Show that 9ⁿ⁺¹ – 8n – 9 is divisible by 64, whenever n is a positive integer.
Solution:
In order to show that 9ⁿ⁺¹ – 8n – 9 is divisible by 64,
it has to be shown that 9ⁿ⁺¹ – 8n – 9 = 64k, wherek k is some natural number
using the binomial theorem

Where, K \(\left[{ }^{n+1} C_2+{ }^{n+1} C_3(8)+\ldots .+{ }^{n+1} C_{n+1}(8)^{n-1}\right]\) is a natural number.
Thus, 9ⁿ⁺¹ – 8n – 9 is divisible by 64 whenever n is a positive integer.
Hence proved.

Question 9.
Prove that \(\sum_{r=0}^n 3^{r n} C_r=4^n\)
Solution:
By the Binomial theorem
\(\sum_{r=0}^n{n}{r} a^{n-r} b^r=(a+b)^n\)
On the right side, we need 4n, so we will put the values,
as, putting b = 3 & a = 1 in the above equation, we get

Hence proved.

Question 10.
Using the binomial theorem, prove that 50ⁿ – 49n – 1 is divisible by 49² for all positive integers n.
Solution:

divisible by (26)² = 676
∴ 54ⁿ + 52n – 1 is divisible by 676 for all positive integers n.