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

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

11th Maths Exercise 5.1 Question 1.

Solution:


Similarly (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
∴ (a – b)⁴ + (a + b)⁴ = 2 [a⁴ + 6a²b² + b⁴]
Substituting the value of a and b we get

= 2[16x⁸ + 6(4x⁴)(9(1 – x²)) + 81(1 – x²)²]
= 2[16x⁸ + 216x⁴(1 – x²) + 81(1 – x²)²]
= 2[16x⁸ + 216x⁴ – 216x⁶ + 81 + 81x⁴ – 162x²]
= 2[16x⁸ – 216x⁶ + 297x⁴ – 162x² + 81]
= 32x⁸ – 432x⁶ + 594x⁴ – 324x² + 162

11th Maths Exercise 5.1 Samacheer Kalvi Question 2.
Compute
(i) 102⁴
(ii) 99⁴
(iii) 9⁷
Solution:

(i) 102⁴ = (100 + 2)⁴ = (10² + 2)⁴

= 1(10⁸) + 4(10⁶)(2) + 6(10⁴)(4) + 4(10²)(8) + 16
= 100000000 + 8000000 + 240000 + 3200 + 16
= 108243216

(ii) 994 = (100 – 1)⁴ = (10² – 1)⁴

= 1(10⁸) + 4(10⁶)(-1) + 6 (10⁴)(1) + 4( 10⁴)(-1) + (-1)⁴
= 100000000 – 4000000 + 60000 – 400 + 1
= 100060001 – 4000400 = 96059601

(iii) 9⁷ = (10 – 1)⁷

= 1(10000000) + 7(1000000)(-1) + 21(100000)(1) + 35(10000)(-1) + 35(1000)(1) + 21(100)(-1) + 7(10)(1) + 1(-1)
= 10000000 – 7000000 + 2100000 – 350000 + 35000 – 2100 + 70 – 1
= 12135070 – 7352101 = 4782969

Sequences And Series Solutions Question 3.
Using binomial theorem, indicate which of the following two number is larger: (1.01)^1000000, 10000.
Solution:
(1.01)^1000000 = (1 + 0.01)^1000000

which is > 10000
So (1.01)^1000000 > 10000 (i.e.) (1.01)^1000000 is larger

11 Maths Samacheer Kalvi Question 4.

Solution:

To find coefficient of x¹⁵ we have to equate x power to 15
i.e. 20 – 5r = 15
20 – 15 = 5r ⇒ 5r = 5 ⇒ r = 5/5 = 1
So the coefficient of x¹⁵ is ¹⁰C₁ = 10

Ex 5.1 Class 11 Question 5.

Solution:

To find coefficient of x⁶
12 – 5r = 6
12 – 6 = 5r ⇒ 5r = 6 ⇒ r = 6/5 which is not an integer.
∴ There is no term involving x⁶.
To find coefficient of x²
12 – 5r = 2
5r = 12 – 2 = 10 ⇒ r = 2

Question 6.

Solution:

when multiplying these terms, we get x⁴ terms

∴ The co-eff of x⁴ is 26325

Ex 5.1 Class 11 Question 7.

Solution:

Chapter 5 Class 11 Maths Question 8.
Find the last two digits of the number 3⁶⁰⁰.
Solution:
3⁶⁰⁰ = 3^{2 × 300} = (9)300 = (10 – 1)³⁰⁰

All the terms except last term are ÷ by 100. So the last two digits will be 01.

Samacheer Kalvi 11th Maths Question 9.
If n is a positive integer, show that, 9^{n + 1} – 8n – 9 is always divisible by 64.
Solution:

∴ 9^{n + 1} – 8n – 9 = 64 [an integer]
⇒ 9^{n + 1} – 8n – 9 is divisible by 64

Class 11 Maths Binomial Theorem 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.
Solution:
Given n is odd. So let n = 2n + 1, where n is an integer.
The expansion (x + y)ⁿ has n + 1 terms.
= 2n + 1 + 1 = 2(n + 1) terms which is an even number.

⇒ The coefficient of the middle terms in (x + y)ⁿ are equal.

Class 11 Ex 5.1 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.
Solution:

Class 11 Maths Chapter 5 Exercise 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. [Hint: write aⁿ = (a – b + b)ⁿ and expand]
Solution:

= (a – b)[aⁿ integer]
⇒ aⁿ – bⁿ is divisible by (a – b)

Chapter 5 Maths Class 11 Question 13.
In the binomial expansion of (a + b)aⁿ, the coefficients of the 4^th and 13^th terms are equal to each other, find n.
Solution:
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]

Class 11 Chapter 5 Maths Question 14.
If the binomial coefficients of three consecutive terms in the expansion of (a + x)ⁿ are in the ratio 1 : 7 : 42, then find n.
Solution:
In (a + x)ⁿ general term is t_{r + 1} = ⁿC_r
So, the coefficient of t_{r + 1} is ⁿC_r
We are given that the coefficients of three consecutive terms are in the ratio 1 : 7 : 42.

Class 11 Maths Chapter 5 Question 15.
In the binomial coefficients of (1 + x)ⁿ, the coefficients of the 5^th, 6^th and 7 terms are in AP. Find all values of n.
Solution:

⇒ (n – 1)(n – 14) = 0
∴ n = 7 , 14

Question 16.

Solution:

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

Question 1.

Solution:
Let T_{r + 1} be the term in which x³² and x^-17 occurs,

(i) Since x³² occurs in this term
∴ Exponent of x = 32
⇒ 60 – 7r = 32 ⇒ 7r = 28
r = 28 ÷ 7 = 4
∴ Coefficient of the term containing x³² = ¹⁵C₄ (-1)⁴ = 1365

(ii) Since x^-17 occurs in this term .
∴ Exponent of x = -17
⇒ 60 – 7r = -17 ⇒ 7r = 77, ∴ r = 11

Question 2.
Find a positive value of m for which the coefficient of x² in the expansion of (1 + x)^m is 6.
Solution:

Also, coefficient of x² in the expansion of (1 + x)^m is 6

⇒ m (m – 1) = 4.3 ⇒ m = 4

Question 3.
In the binomial expansion of (1 + a)^{m + n}, prove that the coefficients of a^m and aⁿ are equal.
Solution:

Question 14.
The coefficient of (r – 1)^th, r^th, and (r + 1)^th terms in the expansion of (x + 1)ⁿ are in the ratio 1 : 3 : 5. Find both n and r.
Solution:
We know that co-effcients of (r – 1)^th, r^th and (r + 1)^th terms in the expansion of (x + 1)ⁿ are ⁿC_{r – 2}: ⁿC_{r – 1} and ⁿC_r respectively

⇒ 3n – 8r + 3 = 0 and n – 4r + 5 = 0
Solving these for n, r we get,
n = 7 and r = 3

Question 5.

Solution:

Using binomial theorem,

Question 6.
Show that the coefficient of the middle term in the expansion of (1 + x)²ⁿ is equal to the sum of the coefficients of the two middle terms in the expansion of (1 + x)^{2n – 1}.
Solution:
In the expansion of (1 + x)²ⁿ,
Number of terms = 2n + 1, which is odd

Question 7.
If three consecutive coefficients in the expansion of (1 + x)ⁿ are in the ratio 6 : 33 : 110, find n.
Solution:
Let the consecutive coefficients ⁿC_r, ⁿC_{r + 1} and ⁿC_{r + 2} be the coefficients of T_{r + 1}, T_{r + 2} and T_{r + 3} then ⁿC_r : ⁿC_{r + 1} : ⁿC_{r + 2} = 6 : 33 : 110

Question 8.
If the sum of the coefficients in the expansion of (x + y)ⁿ is 4096. Then find the greatest coefficient in the expansion.
Solution:
Given that, Sum of the coefficients in the expansion of (x + y)ⁿ = 4096
∴ ⁿC₀ + ⁿC₁ + ⁿC₂+…+ ⁿC_n = 4096
[∴ Sum of binomial coefficients in the expansion of (x + a)ⁿ is 2ⁿ]
⇒ 2ⁿ = 4096 = 2¹²

Question 9.

Solution:
The general term in the expansion of

Question 10.

Solution: