Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1

Question 1.
Expand
(i) \(\left(2 x^{2}-\frac{3}{x}\right)^{3}\)
(ii) \(\left(2 x^{2}-3 \sqrt{1-x^{2}}\right)^{4}+\left(2 x^{2}+3 \sqrt{1-x^{2}}\right)^{4}\)
Answer:
(i) \(\left(2 x^{2}-\frac{3}{x}\right)^{3}\)

(ii) \(\left(2 x^{2}-3 \sqrt{1-x^{2}}\right)^{4}+\left(2 x^{2}+3 \sqrt{1-x^{2}}\right)^{4}\)

Question 2.
Compute
(i) 102⁴
(ii) 99⁴
(iii) 9⁷
Answer:
(i) 102⁴
102⁴ = (100 + 2)⁴

= 100000000 + 8 × 1000000 + 24 × 10000 + 3200 + 16
= 100000000 + 8000000 + 240000 + 3216
= 108243216

(ii) 99⁴
99⁴ = (100 – 1)⁴

= 100000000 – 4 × 1000000 + 6 × 10000 – 400 + 1
= 100000000 – 4000000 + 60000 -400+1
= 96059601

(iii) 9⁷
9⁷ = (10 – 1)⁷


= 10³ (10⁴ – 7 × 10³ + 21 × 10² – 35 × 10 + 35) – 21 × 100 + 70 – 1
= 10³ (10000 – 7000 + 2100 – 350 + 35 ) – 2100 + 70 – 1
= 10³ (12135 – 7350) – 2031
= 10³ × 4785 – 2031
= 4785000 – 2031
= 4782969

Question 3.
Using binomial theorem, indicate which of the following two number is larger. (l.Ol)1000000, 10000 Answer:

= 1000000C₀ + 1000000C₁( . 01) + …………………….
= 1 + 1000000 × 0 . 01 + ……………………..
= 1 + 10000 + other positive terms
= 10001 + other positive terms
(1.01)^1000000 – 10000 = 10001 + other positive terms – 10000 > 0
∴ (1.01)^1000000 > 10000 ⇒ (1.01)^1000000 is larger.

Question 4.
Find the coefficient of x¹⁵ in \(\left(x^{2}+\frac{1}{x^{3}}\right)^{10}\)
Answer:
General term T_r+1 = nC_r x^n-r . a^r
∴ In the expansion \(\left(x^{2}+\frac{1}{x^{3}}\right)^{10}\)

To find the coefficient of x¹⁵, Put 20 – 5r = 15
20 – 15 = 5r ⇒ 5r = 5 ⇒ r = 1
∴ T_{1 + 1} = 10C₁x^{20 – 5} ⇒ T₂ = 10 . x¹⁵
∴ The coefficient of x¹⁵ is 10

Question 5.
Find the coefficient of x6 and the coefficient of x² in \(\left(x^{2}-\frac{1}{x^{3}}\right)^{6}\)
Answer:
The general terms is T_r+1 = nC_ra^n-r . a^r
∴ The general term in the expansion of \(\left(x^{2}-\frac{1}{x^{3}}\right)^{6}\) is

To find the coefficient of x⁶ , Put 12 – 5r = 6
12 – 6 = 5r ⇒ 5r = 6 ⇒ r = \(\frac{6}{5}\) which is impossible.
∴ There is no x⁶ term in the expansion.
To find the coefficient of x²,
Put 12 – 5r = 2
⇒ 12 – 2 = 5r
⇒ 10 = 5r
⇒ r = \(\frac{10}{5}\) = 5
Substituting in (1) we have

T₃ = 15
∴ The coefficient of x² is 15

Question 6.
Find the coefficient of x⁴ in the expansion of (1 + x³)⁵⁰ \(\left(x^{2}+\frac{1}{x}\right)^{5}\)
Answer:


= 10x⁴ × 1 + 5 × 1225 x⁴ + 19600 x⁴ + 500 x⁴
= x⁴ (10 + 6125 + 19600 + 500) = 26235 . x⁴
∴ The coefficient of x⁴ is 26235.

Question 7.
Find the constant term of \(\left(x^{3}-\frac{1}{3 x^{2}}\right)^{5}\)
Answer:


To get the constant term,
Put 15 – 5r = 0
⇒ 5r = 15
⇒ r = 3

Question 8.
Find the last two digits of the number 3⁶⁰⁰.
Answer:
Consider 3⁶⁰⁰
3⁶⁰⁰ = (3²)³⁰⁰ = 9³⁰⁰ = (1o – 1)³⁰⁰

= 10³⁰⁰ – 300 (10)²⁹⁹ + ………………. + 300 C₁ × 10 × – 1 + 1 × 1 × 1
= 10³⁰⁰ – 300 (1o)²⁹⁹ + …………….. – 300 × 10 + 1
= 10³⁰⁰ – 300 × 10²⁹⁹ + ……………… – 3000 + 1
All the terms except the last are multiples of 100 and hence divisible by 100.
∴ The last two digits will be 01.

Question 9.
If n is a positive integer show that 9ⁿ⁺¹ – 8n – 9 is always divisible by 64.
Answer:

which is divisible by 64 for all positive integer n.
∴ 9ⁿ – 8n – 1 is divisible by 64 for all positive integer n.
Put n = n + 1 we get
9^{n + 1} – 8 (n + 1) – 1 is divisible by 64 for all possible integer n
(9^{n + 1} – 8n – 8 – 1) is divisible by 64
∴ 9^{n + 1} – 8n – 9 is always divisible by 64

Question 10.
If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y)ⁿ are equal.
Answer:
Given (x + y)ⁿ
If n is odd the middle term in the expansion of (x + y)ⁿ are \(\frac{T_{n-1}}{2}+1\) and \(\mathrm{T}_{\frac{\mathrm{n}+1}{2}+1}\)

Question 11.
If n is a positive integer and r is a non-negative integer, prove that the coefficients of x^r and x^n-r in the expansion of (1 + x)ⁿ are equal.
Answer:
Given (1 + x)ⁿ.
General term T_r+1 = nC_r x^{n – r} . a^r
∴ The general term in the expansion of (1 + x)ⁿ is T_r+1 = nC_r . (1)^{n – r} . x^r
T_r+1 = nC_r . x^r ……….. (1)
∴ Coefficient of x^r is nC_r,
Put r = n – r in (1)
T_{n – r + 1} = nC_{n – r} . x^n-r
∴ The coefficient of x^n-r is nC_n-r …………. (2)
we know nC_r = nC^n-r
∴ The coefficient of x^r and coefficient of x^n-r are equal, (by (1) & (2))

Question 12.
If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer.
Answer:
Let a = a + b – b
= b + (a – b)
aⁿ = [b + (a – b)]ⁿ
Using binomial expansion

Question 13.
In the binomial expansion of (a + b )n , the coefficients of the 4th and 13th terms are equal to each other , find n.
Answer:
In (a + b)ⁿ general term is t_{r + 1} = ⁿC_r a^{n – r} b^r
So, t₄ = t_{3 + 1} = ⁿC₃ = ⁿC₁₂
⇒ n = 12 + 3 = 15
We are given that their coefficients are equal ⇒ ⁿC₃ = ⁿC₁₂ ⇒ n = 12 + 3 = 15
[ⁿC_x = ⁿC_y ⇒ x = y (or) x + y = n]

Question 14.
If the binomial co – efficients of three consecutive terms in the expansion of (a + x)ⁿ are in the ratio 1 : 7 : 42, then find n.
Answer:
The general term in the expansion of (a + b)ⁿ is T_r+1 = nC_r . a^n-r . b^r
∴ The general term in the expansion of (a + x)ⁿ is T_r+1 = nC_r . a^n-r . x^r
Let the three consecutive terms be r^th term, (r + 1 )^th term, (r + 2 )^th term.

Question 15.
In the binomial coefficients of (1 + x)ⁿ, the coefficients of the 5^th, 6^th, and 7^th terms are in A.P. Find all values of n.
Answer:
Given (1 + x)ⁿ

12(n – 4) = 30 + n² – 5n – 4n + 20 ⇒ 12n – 48 = 30 + n² – 9n + 20
⇒ n² – 9n – 12n + 50 + 48 = 0 ⇒ n² – 21n + 98 = 0
⇒ n² – 14n – 7n + 98 = 0 ⇒ n(n – 14) – 7(n – 14) = 0
⇒ (n – 7) (n – 14) = 0 ⇒ n = 7 or n = 14

Question 16.
prove that

Answer:

This relation is true for all values of n. Equating coefficient of xⁿ on both sides, we have
The general term in the expansion of (1 + x)²ⁿ is T_{r + 1} = 2nC_r (1)^{2n – r} . x^r
Put r = n we get, T_n+1 = 2nC_n . xⁿ
∴ The coefficient of xⁿ in the expansion of (1 + x)²ⁿ is 2nC_n