Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Miscellaneous Exercise 4 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Miscellaneous Exercise 4

(I) Select the correct answers from the given alternatives.

Question 1.

The total number of terms in the expression of (x + y)^{100} + (x – y)^{100} after simplification is:

(A) 50

(B) 51

(C) 100

(D) 202

Answer:

(B) 51

Hint:

Question 2.

The middle term in the expansion of (1 + x)^{2n} will be:

(A) (n – 1)^{th}

(B) n^{th}

(C) (n + 1)^{th}

(D) (n + 2)^{th}

Answer:

(C) (n + 1)^{th}

Hint:

(1 + x)^{2n} has (2n + 1) terms.

∴ (n + 1 )^{th} term is the middle term.

Question 3.

In the expansion of (x^{2} – 2x)^{10}, the coefficient of x^{16} is

(A) -1680

(B) 1680

(C) 3360

(D) 6720

Answer:

(C) 3360

Hint:

(x^{2} – 2x)^{10} = x^{10} (x – 2)^{10}

To get the coefficient of x^{16} in (x^{2} – 2x)^{10},

we need to check coefficient of x^{6} in (x – 2)^{10}

∴ Required coefficient = ^{10}C_{6} (-2)^{4}

= 210 × 16

= 3360

Question 4.

The term not containing x in expansion of \((1-x)^{2}\left(x+\frac{1}{x}\right)^{10}\) is

(A) ^{11}C_{5}

(B) ^{10}C_{5}

(C) ^{10}C_{4}

(D) ^{10}C_{7}

Answer:

(A) ^{11}C_{5}

Hint:

Question 5.

The number of terms in expansion of (4y + x)^{8} – (4y – x)^{8} is

(A) 4

(B) 5

(C) 8

(D) 9

Answer:

(A) 4

Hint:

Question 6.

The value of ^{14}C_{1} + ^{14}C_{3} + ^{14}C_{5} + …. + ^{14}C_{11} is

(A) 2^{14} – 1

(B) 2^{14} – 14

(C) 2^{12}

(D) 2^{13} – 14

Answer:

(D) 2^{13} – 14

Hint:

Question 7.

The value of ^{11}C_{2} + ^{11}C_{4} + ^{11}C_{6} + ^{11}C_{8} is equal to

(A) 2^{10} – 1

(B) 2^{10} – 11

(C) 2^{10} + 12

(D) 2^{10} – 12

Answer:

(D) 2^{10} – 12

Hint:

Question 8.

In the expansion of (3x + 2)^{4}, the coefficient of the middle term is

(A) 36

(B) 54

(C) 81

(D) 216

Answer:

(D) 216

Hint:

(3x + 2)^{4} has 5 terms.

∴ (3x + 2)^{4} has 3rd term as the middle term.

The coefficient of the middle term

= 6 × 9 × 4

= 216

Question 9.

The coefficient of the 8th term in the expansion of (1 + x)^{10} is:

(A) 7

(B) 120

(C) ^{10}C_{8}

(D) 210

Answer:

(B) 120

Hint:

r = 7

t_{8} = ^{10}C_{7} x^{7} = ^{10}C_{3} x^{7}

∴ Coefficient of 8th term = ^{10}C_{3} = 120

Question 10.

If the coefficients of x^{2} and x^{3} in the expansion of (3 + ax)^{9} are the same, then the value of a is

(A) \(-\frac{7}{9}\)

(B) \(-\frac{9}{7}\)

(C) \(\frac{7}{9}\)

(D) \(\frac{9}{7}\)

Answer:

(D) \(\frac{9}{7}\)

Hint:

(II) Answer the following.

Question 1.

Prove by the method of induction, for all n ∈ N.

(i) 8 + 17 + 26 + ….. + (9n – 1) = \(\frac{n}{2}\) (9n + 7)

Solution:

Let P(n) ≡ 8 + 17 + 26 +…..+(9n – 1) = \(\frac{n}{2}\) (9n + 7), for all n ∈ N.

Step I:

Put n = 1

L.H.S. = 8

R.H.S. = \(\frac{1}{2}\) [9(1) + 7] = 8

∴ L.H.S. = R.H.S.

∴ P(n) is true for n = 1.

Step II:

Let us assume that P(n) is true for n = k.

∴ 8 + 17 + 26 +…..+ (9k – 1) = \(\frac{k}{2}\) (9k + 7) ……(i)

Step III:

We have to prove that P(n) is true for n = k + 1,

i.e., 8 + 17 + 26 + …… + [9(k + 1) – 1]

∴ P(n) is true for n = k + 1.

Step IV:

From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ 8 + 17 + 26 +…..+ (9n – 1) = \(\frac{n}{2}\) (9n + 7) for all n ∈ N.

(ii) 1^{2} + 4^{2} + 7^{2} + …… + (3n – 2)^{2} = \(\frac{n}{2}\) (6n^{2} – 3n – 1)

Solution:

Let P(n) = 1^{2} + 4^{2} + 7^{2} + ….. + (3n – 2)^{2} = \(\frac{n}{2}\) (6n^{2} – 3n – 1), for all n ∈ N.

Step I:

Put n = 1

L.H.S.= 1^{2} = 1

R.H.S.= \(\frac{1}{2}\) [6(1)^{2} – 3(1) – 1] = 1

∴ L.H.S. = R.H.S.

∴ P(n) is true for n = 1.

Step II:

Let us assume that P(n) is true for n = k.

∴ 1^{2} + 4^{2} + 7^{2} +…..+ (3k – 2)^{2} = \(\frac{k}{2}\) (6k^{2} – 3k – 1) ……(i)

Step III:

We have to prove that P(n) is true for n = k + 1,

i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:

From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ 1^{2} + 4^{2} + 7^{2} + … + (3n – 2)^{2} = \(\frac{n}{2}\) (6n^{2} – 3n – 1) for all n ∈ N.

(iii) 2 + 3.2 + 4.2^{2} + …… + (n + 1) 2^{n-1} = n . 2^{n}

Solution:

Let P(n) ≡ 2 + 3.2 + 4.2^{2} +…..+ (n + 1) 2^{n-1} = n.2^{n}, for all n ∈ N.

Step I:

Put n = 1

L.H.S. = 2

R.H.S. = 1(2^{1}) = 2

∴ L.H.S. = R.H.S.

∴ P(n) is true for n = 1.

Step II:

Let us assume that P(n) is true for n = k.

∴ 2 + 3.2 + 4.2^{2} + ….. + (k + 1) 2^{k-1} = k.2^{k} …..(i)

Step III:

We have to prove that P(n) is true for n = k + 1,

i.e., to prove that

2 + 3.2 + 4.2^{2} +….+ (k + 2) 2^{k} = (k + 1) 2^{k+1}

∴ P(n) is true for n = k + 1.

Step IV:

From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ 2 + 3.2 + 4.2^{2} +……+ (n + 1) 2^{n-1} = n.2^{n} for all n ∈ N.

(iv) \(\frac{1}{3.4 .5}+\frac{2}{4.5 .6}+\frac{3}{5.6 .7}+\ldots+\frac{n}{(n+2)(n+3)(n+4)}\) = \(\frac{n(n+1)}{6(n+3)(n+4)}\)

Solution:

Question 2.

Given that t_{n+1} = 5t_{n} – 8, t_{1} = 3, prove by method of induction that t_{n} = 5^{n-1} + 2.

Solution:

Let the statement P(n) has L.H.S. a recurrence relation t_{n+1} = 5t_{n} – 8, t_{1} = 3

and R.H.S. a general statement t_{n} = 5^{n-1} + 2.

Step I:

Put n = 1

L.H.S. = 3

R.H.S. = 5^{1-1} + 2 = 1 + 2 = 3

∴ L.H.S. = R.H.S.

∴ P(n) is true for n = 1.

Put n = 2

L.H.S = t_{2} = 5t_{1} – 8 = 5(3) – 8 = 7

R.H.S. = t_{2} = 5^{2-1} + 2 = 5 + 2 = 7

∴ L.H.S. = R.H.S.

∴ P(n) is tme for n = 2.

Step II:

Let us assume that P(n) is true for n = k.

∴ t_{k+1} = 5t_{k} – 8 and t_{k} = 5^{k-1} + 2

Step III:

We have to prove that P(n) is true for n = k + 1,

i.e., to prove that

t_{k+1} = 5^{k+1-1} + 2 = 5^{k} + 2

t_{k+1} = 5t_{k} – 8 and t_{k} = 5^{k-1} + 2 ……[From Step II]

∴ t_{k+1} = 5(5^{k-1} + 2) – 8 = 5^{k} + 2

∴ P(n) is true for n = k + 1.

Step IV:

From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ t_{n} = 5^{n-1} + 2, for all n ∈ N.

Question 3.

Prove by method of induction

\(\left(\begin{array}{cc}

3 & -4 \\

1 & -1

\end{array}\right)^{n}=\left(\begin{array}{cc}

2 n+1 & -4 n \\

n & -2 n+1

\end{array}\right)\), ∀ n ∈ N.

Solution:

Step IV:

From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ \(\left(\begin{array}{cc}

3 & -4 \\

1 & -1

\end{array}\right)^{n}=\left(\begin{array}{cc}

2 n+1 & -4 n \\

n & -2 n+1

\end{array}\right)\), ∀ n ∈ N.

Question 4.

Expand (3x^{2} + 2y)^{5}

Solution:

Here, a = 3x^{2}, b = 2y, n = 5.

Using binomial theorem,

Question 5.

Expand \(\left(\frac{2 x}{3}-\frac{3}{2 x}\right)^{4}\)

Solution:

Question 6.

Find third term in the expansion of \(\left(9 x^{2}-\frac{y^{3}}{6}\right)^{4}\)

Solution:

Question 7.

Find tenth term in the expansion of \(\left(2 x^{2}+\frac{1}{x}\right)^{12}\)

Solution:

Question 8.

Find the middle term(s) in the expansion of

(i) \(\left(\frac{2 a}{3}-\frac{3}{2 a}\right)^{6}\)

Solution:

Here, a = \(\frac{2 a}{3}\), b = \(\frac{-3}{2 a}\), n = 6.

Now, n is even.

∴ \(\frac{\mathrm{n}+2}{2}=\frac{6+2}{2}=4\)

∴ Middle term is t4, for which r = 3.

∴ The Middle term is -20.

(ii) \(\left(x-\frac{1}{2 y}\right)^{10}\)

Solution:

Here, a = x, b = \(-\frac{1}{2 y}\), n = 10.

Now, n is even.

∴ \(\frac{\mathrm{n}+2}{2}=\frac{10+2}{2}=6\)

∴ Middle term is t6, for which r = 5

(iii) (x^{2} + 2y^{2})^{7}

Solution:

Here, a = x^{2}, b = 2y^{2}, n = 7.

Now, n is odd.

∴ \(\frac{\mathrm{n}+1}{2}=\frac{7+1}{2}=4, \frac{\mathrm{n}+3}{2}=\frac{7+3}{2}=5\)

∴ Middle terms are t_{4} and t_{5}, for which r = 3 and r = 4 respectively.

∴ Middle terms are 280x^{8}y^{6} and 560x^{6}y^{8}.

(iv) \(\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{9}\)

Solution:

Question 9.

Find the coefficients of

(i) x^{6} in the expantion of \(\left(3 x^{2}-\frac{1}{3 x}\right)^{9}\)

Solution:

(ii) x^{60} in the expansion of \(\left(\frac{1}{x^{2}}+x^{4}\right)^{18}\)

Solution:

Question 10.

Find the constant term in the expansion of

(i) \(\left(\frac{4 x^{2}}{3}+\frac{3}{2 x}\right)^{9}\)

Solution:

(ii) \(\left(2 x^{2}-\frac{1}{x}\right)^{12}\)

Solution:

Question 11.

Prove by method of induction

(i) log_{a} x^{n} = n log_{a} x, x > 0, n ∈ N

Solution:

(ii) 15^{2n-1} + 1 is divisible by 16, for all n ∈ N.

Solution:

15^{2n-1} + 1 is divisible by 16, if and only if (15^{2n-1} + 1) is is a multiple of 16.

Let P(n) ≡ 15^{2n-1} + 1 = 16m, where m ∈ N.

Step IV:

From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ 15^{2n-1} + 1 is divisible by 16, for all n ∈ N.

(iii) 5^{2n} – 2^{2n} is divisible by 3, for all n ∈ N.

Solution:

Question 12.

If the coefficient of x^{16} in the expansion of (x^{2} + ax)^{10} is 3360, find a.

Solution:

Question 13.

If the middle term in the expansion of \(\left(x+\frac{b}{x}\right)^{6}\) is 160, find b.

Solution:

∴ 160 = \(\frac{6 \times 5 \times 4 \times 3 !}{3 \times 2 \times 1 \times 3 !} \times b^{3}\)

∴ 160 = 20b^{3}

∴ 8 = b^{3}

∴ b = 2

Question 14.

If the coefficients of x^{2} and x^{3} in theexpansion of (3 + kx)^{9} are equal, find k.

Solution:

Question 15.

If the constant term in the expansion of \(\left(x^{3}+\frac{\mathrm{k}}{x^{8}}\right)^{11}\) is 1320, find k.

Solution:

Question 16.

Show that there is no term containing x^{6} in the expansion of \(\left(x^{2}-\frac{3}{x}\right)^{11}\).

Solution:

Question 17.

Show that there is no constant term in the expansion of \(\left(2 x-\frac{x^{2}}{4}\right)^{9}\)

Solution:

Question 18.

State, first four terms in the expansion of \(\left(1-\frac{2 x}{3}\right)^{-1 / 2}\)

Solution:

Question 19.

State, first four terms in the expansion of \((1-x)^{-1 / 4}\).

Solution:

Question 20.

State, first three terms in the expansion of \((5+4 x)^{-1 / 2}\)

Solution:

Question 21.

Using the binomial theorem, find the value of \(\sqrt[3]{995}\) upto four places of decimals.

Solution:

Question 22.

Find approximate value of \(\frac{1}{4.08}\) upto four places of decimals.

Solution:

Question 23.

Find the term independent of x in the expansion of (1 – x^{2}) \(\left(x+\frac{2}{x}\right)^{6}\).

Solution:

Question 24.

(a + bx) (1 – x)^{6} = 3 – 20x + cx^{2} + …, then find a, b, c.

Solution:

Question 25.

The 3rd term of (1 + x)^{n} is 36x^{2}. Find 5th term.

Solution:

Question 26.

Suppose (1 + kx)^{n} = 1 – 12x + 60x^{2} – …… find k and n.

Solution: