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) 102^{4}

(ii) 99^{4}

(iii) 9^{7}

Answer:

(i) 102^{4}

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

= 100000000 + 8 × 1000000 + 24 × 10000 + 3200 + 16

= 100000000 + 8000000 + 240000 + 3216

= 108243216

(ii) 99^{4}

99^{4} = (100 – 1)^{4}

= 100000000 – 4 × 1000000 + 6 × 10000 – 400 + 1

= 100000000 – 4000000 + 60000 -400+1

= 96059601

(iii) 9^{7}

9^{7} = (10 – 1)^{7}

= 10^{3} (10^{4} – 7 × 10^{3} + 21 × 10^{2} – 35 × 10 + 35) – 21 × 100 + 70 – 1

= 10^{3} (10000 – 7000 + 2100 – 350 + 35 ) – 2100 + 70 – 1

= 10^{3} (12135 – 7350) – 2031

= 10^{3} × 4785 – 2031

= 4785000 – 2031

= 4782969

Question 3.

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

= 1000000C_{0} + 1000000C_{1}( . 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 x^{15} 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 x^{15}, Put 20 – 5r = 15

20 – 15 = 5r ⇒ 5r = 5 ⇒ r = 1

∴ T_{1 + 1} = 10C_{1}x^{20 – 5} ⇒ T_{2} = 10 . x^{15}

∴ The coefficient of x^{15} is 10

Question 5.

Find the coefficient of x6 and the coefficient of x^{2} in \(\left(x^{2}-\frac{1}{x^{3}}\right)^{6}\)

Answer:

The general terms is T_{r+1} = nC_{r}a^{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 x^{6} , Put 12 – 5r = 6

12 – 6 = 5r ⇒ 5r = 6 ⇒ r = \(\frac{6}{5}\) which is impossible.

∴ There is no x^{6} term in the expansion.

To find the coefficient of x^{2},

Put 12 – 5r = 2

⇒ 12 – 2 = 5r

⇒ 10 = 5r

⇒ r = \(\frac{10}{5}\) = 5

Substituting in (1) we have

T_{3} = 15

∴ The coefficient of x^{2} is 15

Question 6.

Find the coefficient of x^{4} in the expansion of (1 + x^{3})^{50} \(\left(x^{2}+\frac{1}{x}\right)^{5}\)

Answer:

= 10x^{4} × 1 + 5 × 1225 x^{4} + 19600 x^{4} + 500 x^{4}

= x^{4} (10 + 6125 + 19600 + 500) = 26235 . x^{4}

∴ The coefficient of x^{4} 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 3^{600}.

Answer:

Consider 3^{600}

3^{600} = (3^{2})^{300} = 9^{300} = (1o – 1)^{300}

= 10^{300} – 300 (10)^{299} + ………………. + 300 C_{1} × 10 × – 1 + 1 × 1 × 1

= 10^{300} – 300 (1o)^{299} + …………….. – 300 × 10 + 1

= 10^{300} – 300 × 10^{299} + ……………… – 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 9^{n+1} – 8n – 9 is always divisible by 64.

Answer:

which is divisible by 64 for all positive integer n.

∴ 9^{n} – 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 nonnegative 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 a^{n} – b^{n}, whenever n is a positive integer.

Answer:

Let a = a + b – b

= b + (a – b)

a^{n} = [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 the expansion of (a + b)^{n},

The general term is T_{r+1} = nC_{r} . a^{n-r} . b^{r} ……….. (1)

To find the coefficient of 4th term, Put r = 3 in equation (1)

∴ T_{3+1} = nC_{3} a^{n-3} . b^{3}

To find the coefficient of 13th term , Put r = 12 in equation (1)

∴ T_{12+1} = nC_{12} a^{n-12} . b^{12}

Given nC_{3} = nC_{12}

nC_{x} = nC_{y} ⇒ x = y or x + y = n

∴ 3 + 12 = n ⇒ n = 15

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 + n^{2} – 5n – 4n + 20 ⇒ 12n – 48 = 30 + n^{2} – 9n + 20

⇒ n^{2} – 9n – 12n + 50 + 48 = 0 ⇒ n^{2} – 21n + 98 = 0

⇒ n^{2} – 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 x^{n} 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} . x^{n}

∴ The coefficient of x^{n} in the expansion of (1 + x)^{2n} is 2nC_{n}