Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(a) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Exercise 9(a)

Question 1.

The rows n = 6 and n = 7 in the pascal triangle have been kept vacant. Fill in the gaps

Solution:

Question 2.

Write down the expansion of (a + b)^{8} using Pascal’s triangle.

Solution:

The row n = 8 in Pascal’s triangle is 1, 8, 28, 56, 70, 56, 28, 8, 1.

∴ (a + b)^{8} = a^{8} + 8a^{8-1}b^{1} + 28a^{8-2}b^{2} + 56a^{8-3}b^{3 }+ 70a^{8-4}b^{4} + 56a^{8-5}b^{5} + 28a^{8-6}b^{6} + 8a^{8-7}b^{7} + b^{8}

= a^{8} + 8a^{7}b + 28a^{6}b^{2} + 56a^{5}b^{3} + 70a^{4}b^{4} + 56a^{3}b^{5} + 28a^{2}b^{6} + 8ab^{7} + b^{8}

Question 3.

Find the 3rd term in the expansion of \(\left(2 x^3-\frac{1}{x^6}\right)^4\) using rules of Pascal triangle.

Solution:

Question 4.

Expand the following :

(a) (7a + 3b)^{6}

Solution:

(7a + 3b)^{6} = (7a)^{6} + ^{6}C_{1}(7a)^{6-1}(3b)^{1} + ^{6}C_{2}(7a)^{6-2}(3b)^{2} + ….. + (3b)^{6}7^{6}a^{6} + 6(7a)^{5}(3b) + 15(7a^{4}) × 9b^{2} + …. + 3^{6}b^{6}

= 7a^{6} + 18 × 7^{5}a^{5}b + 135 × 7^{4}a^{4}b^{2} + ….. + 3^{6}b^{6}

(b) \(\left(\frac{-9}{2} a+b\right)^7\)

Solution:

(c) \(\left(a-\frac{7}{3} c\right)^4\)

Solution:

Question 5.

Apply Binominal Theorem to find the value of (1.01)^{5}.

Solution:

= 1 + ^{5}C_{1}(0.01)^{1} + ^{5}C_{2}(0.01)^{2} + ^{5}C_{3}(0.01)^{3} + ^{5}C_{4}(0.01)^{4 }+ (0.01)^{5}

= 1 + 5 × 0.01 + 10(0.0001) + 10(0.000001) + 5(0.00000001) + 0.0000000001

= 1 + 0.05 + 0.001 + 0.00001 + 0.00000005 + 0.000000001

= 1.0510100501

Question 6.

State true or false.

(a) The number of terms in the expansion of \(\left(x^2-2+\frac{1}{x^2}\right)^6\) is equal to 7.

Solution:

False

(b) There is a term independent of both x and y in the expansion of \(\left(x^2+\frac{1}{y^2}\right)^9\)

Solution:

False

(c) The highest power in the expansion of \(x^{40}\left(x^2+\frac{1}{x^2}\right)^{20}\) is equal to 40.

Solution:

False

(d) The product of K consecutive natural numbers is divisible by K!

Solution:

True

Question 7.

Answer the following :

(a) If the 6th term in the expansion of (x + *)^{n} is equal to ^{n}C_{5}x^{n-10} find *

Solution:

Let the 6th term (x + y)^{n} is ^{n}C_{5}x^{n-10}

∴ ^{n}C_{5}x^{n-5}y^{5} = ^{n}C_{5}x^{n-10} = ^{n}C_{5}x^{n-5}.x^{-5}

⇒ y^{5} = x^{-5} = \(\frac{1}{x^5}\)

∴ y = \(\frac{1}{x}\) . Hence * = \(\frac{1}{x}\)

(b) Find the number of terms in the expansion of (1 + x)^{n} (1 – x)^{n}.

Solution:

(1 + x)^{n} (1 – x)^{n} = (1 – x^{2})^{n}

∴ The number of terms in this expansion is (n + 1)

(c) Find the value of \(\frac{{ }^n \mathrm{C}_{r-1}}{{ }^n \mathrm{C}_r}\)

Solution:

(d) How many terms in the expansion of \(\left(\frac{3}{a}+\frac{a}{3}\right)^{10}\) have positive powers of a? How many have negative powers of a?

Solution:

Question 8.

Find the middle term(s) in the expansion of the following.

(a) \(\left(\frac{a}{b}+\frac{b}{a}\right)^6\)

Solution:

Here there is only one middle term i.e. the 4th term.

∴ 4th term i.e. (3 + 1)th term in the expansion of

(b) \(\left(x+\frac{1}{x}\right)^9\)

Solution:

Here there are two middle terms i.e. 5th and 6th terms.

∴ 5th term in the above expansion is

(c) \(\left(x^{\frac{3}{2}}-y^{\frac{3}{2}}\right)^8\)

Solution:

Here there is only one middle term i.e. 5th term.

∴ 5th term i.e. (4 + 1)th term in the expansion of

Question 9.

Find the 6th term in the expansion of \(\left(x^2+\frac{a^4}{y^2}\right)^{10}\)

Solution:

6th term i.e. (5 + 1)th term in the expansion of

(x^{2} + \(\frac{a^4}{y^2}\))^{10} is ^{10}C_{5}(x^{2})^{10-5} (\(\frac{a^4}{y^2}\))^{5}

Question 10.

(a) Find the fifth term in the expansion of (6x – \(\frac{a^3}{x}\))^{10}

Solution:

(b) Is there a term independent of x? If yes find it out.

Solution:

Let (r + 1)th term in the expansion of (6x – \(\frac{a^3}{x}\))^{10} is independent of x.

∴ (r + 1)th term = ^{10}C_{r}(6x)^{10-r}(\(\frac{-a^3}{x}\))^{r}

= ^{10}C_{r}6^{10-r}x^{10-r}(-1)^{r}a^{3r}x^{-r}

= (-1)^{r 10}C_{r}6^{10-r}a^{3r}x^{10-2r}

∴ x^{10-2r} = 1 = x^{0}

or, 10 – 2r = 0 or, r = 5

∴ 6th term is term independent of x in the above expansion and the term is (-1)^{5} ^{10}C_{5}6^{10-5}a^{3.5}

= – ^{10}C_{5}6^{5}a^{15} = – 252 × 6^{5}a^{15}

Question 11.

(a) Find the coefficient of \(\frac{1}{y^{10}}\) in the expansion of \(\left(y^3+\frac{a^7}{y^5}\right)^{10}\)

Solution:

(b) Does there exist a term independent of y in the above expansion?

Solution:

Let (r + 1)th term is independent of y.

∴ y^{30-8r} = 1 = y^{0} or, 30 – 8r = 0

or, r = \(\frac{30}{8}\) = \(\frac{15}{4}\) which is not possible as r∈ N or zero.

∴ There is no term in the expansion of \(\left(y^3+\frac{a^7}{y^5}\right)^{10}\) which is independent of y.

Question 12.

(a) Find the coefficient of x^{4} in the expansion of (1 + 3x + 10x^{2})(x + \(\frac{1}{x}\))^{10}

Solution:

(b) Find the term independent of x in the above expansion.

Solution:

Question 13.

Show that the coefficient of a^{m} and a^{n} in expansion of (1 + a)^{m+n} are equal.

Solution:

(m + 1)th and (n + 1)th terms in the expansion of (1 + a)^{m+n} are ^{m+n}C_{m}a^{m} and ^{m+n}C_{n}a^{n}

∴ The coefficient of a^{m} and a^{n} are ^{m+n}C_{m} and ^{m+n}C_{n} which are equal.

Question 14.

An expression of the form (a + b + c + d + …. ) consisting of a sum of many distinct symbols called a multinomial. Show that (a + b + c)^{n} is

the sum of all terms of the form \(\frac{\boldsymbol{n} !}{\boldsymbol{p} ! \boldsymbol{q} ! \boldsymbol{r} !}\) a^{p}b^{q}c^{r} where p, q and r range over all positive triples of non-negative integers such that p + q + r = n.

Solution:

Question 15.

State and prove a multinominal Theorem.

Solution:

Multinominal Theorem:

(P_{1} + P_{2} + ……. + P_{m})^{n}

\(=\sum \frac{n !}{n_{1} ! n_{2} ! \ldots n_{m} !} p_1^{n_1} p_2^{n_2} \ldots p_m^{n_m}\)

where n_{1} + n_{2} + ……. + n_{m} = n

The proof of this theorem is beyond the syllabus.