Exercise 2.6 - Chapter 2 - Algebra - 11th Business Maths Guide Samacheer Kalvi Solutions

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Exercise 2.6

Question 1.
Expand the following by using binomial theorem:
(i) (2a – 3b)⁴

Solution:

Question 2.
Evaluate the following using binomial theorem:
(i) (101)⁴
(ii) (999)⁵
Solution:
(i) (x + a)ⁿ = nC₀ xⁿ a⁰ + nC₁ x^n-1 a¹ + nC₂ x^n-2 a² + ……… + nC_r x^n-r a^r + …… + nC_n aⁿ
(101)⁴ = (100 + 1)⁴ = 4C₀ (100)⁴ + 4C₁ (100)³ (1)¹ + 4C₂ (100)² (1)² + 4C₃ (100)¹ (1)³ + 4C₄ (1)⁴
= 1 × (100000000) + 4 × (1000000) + 6 × (10000) + 4 × 100 + 1 × 1
= 100000000 + 4000000 + 60000 + 400 + 1
= 10,40,60,401

(ii) (x + a)ⁿ = nC₀ xⁿ a⁰ + nC₁ x^n-1 a¹ + nC₂ x^n-2 a² + ……… + nC_r x^n-r a^r + …… + nC_n aⁿ
(999)⁵ = (1000 – 1)⁵ = 5C₀ (1000)⁵ – 5C₁ (1000)⁴ (1)¹ + 5C₂ (1000)³ (1)² – 5C₃ (1000)² (1)³ + 5C₄ (1000)⁵ (1)⁴ – 5C₅ (1)⁵
= 1(1000)⁵ – 5(1000)⁴ – 10(1000)³ – 10(1000)² + 5(1000) – 1
= 1000000000000000 – 5000000000000 + 10000000000 – 10000000 + 5000 – 1
= 995009990004999

Question 3.
Find the 5th term in the expansion of (x – 2y)¹³.
Solution:
General term is t_r+1 = nC_r x^n-r a^r
(x – 2y)¹³ = (x + (-2y))¹³
Here x is x, a is (-2y) and n = 13
5th term = t₅ = t₄₊₁ = 13C₄ x^13-4 (-2y)⁴
= 13C₄ x⁹ 2⁴ y⁴

= 13 × 11 × 10 × 8x⁹y⁴
= 13 × 880x⁹y⁴
= 11440x⁹y⁴

Question 4.
Find the middle terms in the expansion of
(i) (x+1x)11
(ii) (3x+x22)8
(iii) (2x2−3x3)10
Solution:
(i) General term is t_r+1 = nC_r x^n-r a^r

Question 5.
Find the term in dependent of x in the expansion of

Solution:
(i) Let the independent form of x occurs in the general term, t_r+1 = nC_r x^n-r a^r

ndependent term occurs only when x power is zero.
18 – 3r = 0
⇒ 18 = 3r
⇒ r = 6
Put r = 6 in (1) we get the independent term as

[∵ 9C₆ = 9C_9-6 = 9C₃]

Let the independent term of x occurs in the general term

Independent term occurs only when x power is zero.
15 – 3r = 0
15 = 3r
r = 5
Using r = 5 in (1) we get the independent term
= 15C₅ x⁰ (-2)⁵ [∵ (-2)⁵ = (-1)⁵ 2⁵ = -2⁵]
= -32(15C₅)

Independent term occurs only when x power is zero
24 – 3r = 0
24 = 3r
r = 8
Put r = 8 in (1) we get the independent term as
= 12C₈ 2^12-8 x⁰
= 12C₄ × 2⁴ × 1
= 7920

Question 6.
Prove that the term independent of x in the expansion of

Solution:
There are (2n + 1) terms in expansion.
∴ t_n+1 is the middle term.

Question 7.
Show that the middle term in the expansion of is 

Solution:
There are 2n + 1 terms in expansion of (1 + x)²ⁿ.
∴ The middle term is t_n+1.