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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} = a^{4} + 4a^{3}b + 6a^{2}b^{2} + 4ab^{3} + b^{4}

∴ (a – b)^{4} + (a + b)^{4} = 2 [a^{4} + 6a^{2}b^{2} + b^{4}]

Substituting the value of a and b we get

= 2[16x^{8} + 6(4x^{4})(9(1 – x^{2})) + 81(1 – x^{2})^{2}]

= 2[16x^{8} + 216x^{4}(1 – x^{2}) + 81(1 – x^{2})^{2}]

= 2[16x^{8} + 216x^{4} – 216x^{6} + 81 + 81x^{4} – 162x^{2}]

= 2[16x^{8} – 216x^{6} + 297x^{4} – 162x^{2} + 81]

= 32x^{8} – 432x^{6} + 594x^{4} – 324x^{2} + 162

**11th Maths Exercise 5.1 Samacheer Kalvi Question 2.**

Compute

(i) 102^{4}

(ii) 99^{4}

(iii) 9^{7}

Solution:

(i) 102^{4} = (100 + 2)^{4} = (10^{2} + 2)^{4}

= 1(10^{8}) + 4(10^{6})(2) + 6(10^{4})(4) + 4(10^{2})(8) + 16

= 100000000 + 8000000 + 240000 + 3200 + 16

= 108243216

(ii) 994 = (100 – 1)^{4} = (10^{2} – 1)^{4}

= 1(10^{8}) + 4(10^{6})(-1) + 6 (10^{4})(1) + 4( 10^{4})(-1) + (-1)^{4}

= 100000000 – 4000000 + 60000 – 400 + 1

= 100060001 – 4000400 = 96059601

(iii) 9^{7} = (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 x^{15} 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 x^{15} is ^{10}C_{1} = 10

**Ex 5.1 Class 11 Question 5.**

Solution:

To find coefficient of x^{6}

12 – 5r = 6

12 – 6 = 5r ⇒ 5r = 6 ⇒ r = 6/5 which is not an integer.

∴ There is no term involving x^{6}.

To find coefficient of x^{2}

12 – 5r = 2

5r = 12 – 2 = 10 ⇒ r = 2

Question 6.

Solution:

when multiplying these terms, we get x^{4} terms

∴ The co-eff of x^{4} 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 3^{600}.

Solution:

3^{600} = 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 a^{n} – b^{n}, whenever n is a positive integer. [Hint: write a^{n} = (a – b + b)^{n} and expand]

Solution:

= (a – b)[a^{n} integer]

⇒ a^{n} – b^{n} is divisible by (a – b)

**Chapter 5 Maths Class 11 Question 13.**

In the binomial expansion of (a + b)a^{n}, 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} = ^{n}C_{r }a^{n – r }b^{r}

So, t_{4} = t_{3 + 1} = ^{n}C_{3} = ^{n}C_{12}

⇒ n = 12 + 3 = 15

We are given that their coefficients are equal ⇒ ^{n}C_{3} = ^{n}C_{12} ⇒ n = 12 + 3 = 15

[^{n}C_{x} = ^{n}C_{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} = ^{n}C_{r}

So, the coefficient of t_{r + 1} is ^{n}C_{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 x^{32} and x^{-17} occurs,

(i) Since x^{32} occurs in this term

∴ Exponent of x = 32

⇒ 60 – 7r = 32 ⇒ 7r = 28

r = 28 ÷ 7 = 4

∴ Coefficient of the term containing x^{32} = ^{15}C_{4} (-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 x^{2} in the expansion of (1 + x)^{m} is 6.

Solution:

Also, coefficient of x^{2} 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 a^{n} 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 ^{n}C_{r – 2}: ^{n}C_{r – 1} and ^{n}C_{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 ^{n}C_{r}, ^{n}C_{r + 1} and ^{n}C_{r + 2} be the coefficients of T_{r + 1}, T_{r + 2} and T_{r + 3} then ^{n}C_{r} : ^{n}C_{r + 1} : ^{n}C_{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

∴ ^{n}C_{0} + ^{n}C_{1} + ^{n}C_{2}+…+ ^{n}C_{n} = 4096

[∴ Sum of binomial coefficients in the expansion of (x + a)^{n} is 2^{n}]

⇒ 2^{n} = 4096 = 2^{12}

Question 9.

Solution:

The general term in the expansion of

Question 10.

Solution: