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) \(\left(2 x^{2}-\frac{3}{x}\right)^{3}\)
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 1

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

(ii) \(\left(2 x^{2}-3 \sqrt{1-x^{2}}\right)^{4}+\left(2 x^{2}+3 \sqrt{1-x^{2}}\right)^{4}\)
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 2
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 3

Question 2.
Compute
(i) 1024
(ii) 994
(iii) 97
Answer:
(i) 1024
1024 = (100 + 2)4
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 4
= 100000000 + 8 × 1000000 + 24 × 10000 + 3200 + 16
= 100000000 + 8000000 + 240000 + 3216
= 108243216

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

(ii) 994
994 = (100 – 1)4
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 5
= 100000000 – 4 × 1000000 + 6 × 10000 – 400 + 1
= 100000000 – 4000000 + 60000 -400+1
= 96059601

(iii) 97
97 = (10 – 1)7
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 6
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 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:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 8
= 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.

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

Question 4.
Find the coefficient of x15 in \(\left(x^{2}+\frac{1}{x^{3}}\right)^{10}\)
Answer:
General term Tr+1 = nCr xn-r . ar
∴ In the expansion \(\left(x^{2}+\frac{1}{x^{3}}\right)^{10}\)
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 9
To find the coefficient of x15, Put 20 – 5r = 15
20 – 15 = 5r ⇒ 5r = 5 ⇒ r = 1
∴ T1 + 1 = 10C1x20 – 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 Tr+1 = nCran-r . ar
∴ The general term in the expansion of \(\left(x^{2}-\frac{1}{x^{3}}\right)^{6}\) is
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 10
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
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 11
T3 = 15
∴ The coefficient of x2 is 15

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

Question 6.
Find the coefficient of x4 in the expansion of (1 + x3)50 \(\left(x^{2}+\frac{1}{x}\right)^{5}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 12
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 13
= 10x4 × 1 + 5 × 1225 x4 + 19600 x4 + 500 x4
= x4 (10 + 6125 + 19600 + 500) = 26235 . x4
∴ The coefficient of x4 is 26235.

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

Question 7.
Find the constant term of \(\left(x^{3}-\frac{1}{3 x^{2}}\right)^{5}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 14
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 15
To get the constant term,
Put 15 – 5r = 0
⇒ 5r = 15
⇒ r = 3
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 16

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

Question 8.
Find the last two digits of the number 3600.
Answer:
Consider 3600
3600 = (32)300 = 9300 = (1o – 1)300
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 17
= 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:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 18
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
9n + 1 – 8 (n + 1) – 1 is divisible by 64 for all possible integer n
(9n + 1 – 8n – 8 – 1) is divisible by 64
∴ 9n + 1 – 8n – 9 is always divisible by 64

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

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 \(\mathrm{T}_{\frac{\mathrm{n}+1}{2}+1}\)
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 19

Question 11.
If n is a positive integer and r is a nonnegative integer, prove that the coefficients of xrand
xn-r in the expansion of (1 + x)n are equal.
Answer:
Given (1 + x)n.
General term Tr+1 = nCr xn – r . ar
∴ The general term in the expansion of (1 + x)n is Tr+1 = nCr . (1)n – r . xr
Tr+1 = nCr . xr ……….. (1)
∴ Coefficient of xr is nCr,
Put r = n – r in (1)
Tn – r + 1 = nCn – r . xn-r
∴ The coefficient of xn-r is nCn-r …………. (2)
we know nCr = nCn-r
∴ The coefficient of xr and coefficient of xn-r are equal, (by (1) & (2))

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

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
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 20

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 the expansion of (a + b)n,
The general term is Tr+1 = nCr . an-r . br ……….. (1)
To find the coefficient of 4th term, Put r = 3 in equation (1)
∴ T3+1 = nC3 an-3 . b3
To find the coefficient of 13th term , Put r = 12 in equation (1)
∴ T12+1 = nC12 an-12 . b12
Given nC3 = nC12
nCx = nCy ⇒ x = y or x + y = n
∴ 3 + 12 = n ⇒ n = 15

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

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 Tr+1 = nCr . an-r . br
∴ The general term in the expansion of (a + x)n is Tr+1 = nCr . an-r . xr
Let the three consecutive terms be rth term, (r + 1 )th term, (r + 2 )th term.
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 21
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 22

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

Question 15.
In the binomial coefficients of (1 + x)n , the coefficients of the 5th, 6th and 7th terms are in A.P. Find all values of n.
Answer:
Given (1 + x)n
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 23
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
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 24
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 25

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

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 Tr + 1 = 2nCr (1)2n – r . xr
Put r = n we get, Tn+1 = 2nCn . xn
∴ The coefficient of xn in the expansion of (1 + x)2n is 2nCn
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 26

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top