Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.1 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.1

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

Question 1.

2 + 4 + 6 + …… + 2n = n(n + 1)

Solution:

Let P(n) = 2 + 4 + 6 + …… + 2n = n(n + 1), for all n ∈ N.

Step I:

Put n = 1

L.H.S. = 2

R.H.S. = 1(1 + 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 + 4 + 6 + ….. + 2k = k(k + 1) ……(i)

Step III:

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

i.e., to prove that

2 + 4 + 6 + …… + 2(k + 1) = (k + 1) (k + 2)

L.H.S. = 2 + 4 + 6 + …+ 2(k + 1)

= 2 + 4 + 6+ ….. + 2k + 2(k + 1)

= k(k + 1) + 2(k + 1) …..[From (i)]

= (k + 1).(k + 2)

= R.H.S.

∴ 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 + 4 + 6 + …… + 2n = n(n + 1) for all n ∈ N.

Question 2.

3 + 7 + 11 + ……… to n terms = n(2n + 1)

Solution:

Let P(n) = 3 + 7 + 11 + ……… to n terms = n(2n +1), for all n ∈ N.

But 3, 7, 11, …. are in A.P.

∴ a = 3 and d = 4

Let t_{n} be the nth term.

∴ t_{n} = a + (n – 1)d = 3 + (n – 1)4 = 4n – 1

∴ P(n) = 3 + 7 + 11 + ……. + (4n – 1) = n(2n + 1)

Step I:

Put n = 1

L.H.S. = 3

R.H.S. = 1[2(1)+ 1] = 3

∴ 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.

∴ 3 + 7 + 11 + ….. + (4k – 1) = k(2k + 1) …..(i)

Sept III:

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

i.e., to prove that

3 + 7 + 11 + …+ [4(k + 1) – 1] = (k + 1)(2k + 3)

L.H.S. = 3 + 7 + 11 + …… + [4(k + 1) – 1]

= 3 + 7 + 11 + ….. + (4k – 1) + [4(k+ 1) – 1]

= k(2k + 1) + (4k + 4 – 1) …..[From (i)]

= 2k^{2} + k + 4k + 3

= 2k^{2} + 2k + 3k + 3

= 2k(k + 1) + 3(k + 1)

= (k + 1) (2k + 3)

= R.H.S.

∴ 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.

∴ 3 + 7 + 11 + ….. to n terms = n(2n + 1) for all n ∈ N.

Question 3.

1^{2} + 2^{2} + 3^{2} +…..+ n^{2} = \(\frac{n(n+1)(2 n+1)}{6}\)

Solution:

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

Step I:

Put n = 1

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

RHS = \(\frac{1(1+1)[2(1)+1]}{6}=\frac{6}{6}\) = 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} + 2^{2} + 3^{2} +…+ k^{2} = \(\frac{k(k+1)(2 k+1)}{6}\) …..(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} + 2^{2} + 3^{2} + …+ n^{2} = \(\frac{n(n+1)(2 n+1)}{6}\) for all n ∈ N.

Question 4.

1^{2} + 3^{2} + 5^{2} + ….. + (2n – 1)^{2} = \(\frac{n}{3}\) (2n – 1)(2n + 1)

Solution:

Let P(n) = 1^{2} + 3^{2} + 5^{2}+…..+ (2n – 1)^{2} = \(\frac{n}{3}\) (2n – 1)(2n + 1), for all n ∈ N.

Step I:

Put n = 1

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

R.H.S. = \(\frac{1}{3}\) [2(1) – 1][2(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} + 3^{2} + 5^{2} +….+(2k – 1)^{2} = \(\frac{k}{3}\) (2k – 1)(2k + 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} + 3^{2} + 5^{2} + …+ (2n – 1)^{2} = \(\frac{n}{3}\) (2n – 1)(2n + 1) for all n ∈ N.

Question 5.

1^{3} + 3^{3} + 5^{3} + ….. to n terms = n^{2} (2n^{2} – 1)

Solution:

Let P(n) = 1^{3} + 3^{3} + 5^{3} + …. to n terms = n^{2} (2n^{2} – 1), for all n ∈ N.

But 1, 3, 5, are in A.P.

∴ a = 1, d = 2

Let tn be the nth term.

t_{n} = a + (n – 1) d = 1 + (n – 1) 2 = 2n – 1

∴ P(n) = 1^{3} + 3^{3} + 5^{3} +…..+ (2n – 1)^{3} = n^{2} (2n^{2} – 1)

Step I:

Put n = 1

L.H.S. = 1^{3} = 1

R.H.S. = 1^{2} [2(1)^{2} – 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^{3} + 3^{3} + 5^{3} +…+ (2k – 1)^{3} = k^{2} (2k^{2} – 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^{3} + 3^{3} + 5^{3} + … to n terms = n^{2} (2n^{2} – 1) for all n ∈ N.

Question 6.

1.2 + 2.3 + 3.4 +… + n(n + 1) = \(\frac{n}{3}\) (n + 1)(n + 2)

Solution:

Let P(n) = 1.2 + 2.3 + 3.4 +….+n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\), for all n ∈ N.

Step I:

Put n = 1

L.H.S. = 1.2 = 2

R.H.S. = \(\frac{1}{3}\) (1 + 1)(1 + 2) = 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.

∴ 1.2 + 2.3 + 3.4 + ….. + k(k + 1) = \(\frac{k}{3}\) (k + 1)(k + 2) ……(i)

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 + 2.3 + 3.4 + … + n(n + 1) = \(\frac{n}{3}\) (n + 1)(n + 2), for all n ∈ N.

Question 7.

1.3 + 3.5 + 5.7 +… to n terms = \(\frac{n}{3}\) (4n^{2} + 6n – 1)

Solution:

Let P(n) = 1.3 + 3.5 + 5.7 +… to n terms = \(\frac{n}{3}\) (4n^{2} + 6n -1), for all n ∈ N.

But first factor in each term, i.e., 1, 3, 5,… are in A.P. with a = 1 and d = 2.

∴ nth term = a + (n – 1)d = 1 + (n – 1) 2 = (2n – 1)

Also, second factor in each term,

i.e., 3, 5, 7, … are in A.P. with a = 3 and d = 2.

∴ nth term = a + (n – 1) d = 3 + (n – 1) 2 = (2n + 1)

∴ nth term, t_{n} = (2n – 1) (2n + 1)

∴ P(n) ≡ 1.3 + 3.5 + 5.7 + …. + (2n – 1) (2n + 1) = \(\frac{n}{3}\) (4n^{2} + 6n – 1)

Step I:

Put n = 1

L.H.S. = 1.3 = 3

R.H.S. = \(\frac{1}{3}\) [4(1)^{2} + 6(1) – 1] = 3

∴ L.H.S. = R.H.S.

∴ P(n) is trae for n = 1.

Step II:

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

∴ 1.3 + 3.5 + 5.7 +….+ (2k – 1)(2k + 1) = \(\frac{k}{3}\) (4k^{2} + 6k – 1) ……(i)

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.

∴ 1.3 + 3.5 + 5.7 +… to n terms = \(\frac{n}{3}\) (4n^{2} + 6n – 1) for all n ∈ N.

Question 8.

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\)

Solution:

Let P(n) ≡ \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\), for all n ∈ N.

Step I:

Put n = 1

L.H.S. = \(\frac{1}{1.3}=\frac{1}{3}\)

R.H.S. = \(\frac{1}{2(1)+1}=\frac{1}{3}\)

∴ 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.

∴ \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 k-1)(2 k+1)}=\frac{k}{2 k+1}\) …..(i)

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.

∴ \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\), for all n ∈ N.

Question 9.

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots \text { to } n \text { terms }=\frac{n}{3(2 n+3)}\)

Solution:

Let P(n) ≡ \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots \text { to } n \text { terms }=\frac{n}{3(2 n+3)}\), for all n ∈ N.

But first factor in each term of the denominator,

i.e., 3, 5, 7, ….. are in A.P. with a = 3 and d = 2.

∴ nth term = a + (n – 1)d = 3 + (n – 1) 2 = (2n + 1)

Also, second factor in each term of the denominator,

i.e., 5, 7, 9, … are in A.P. with a = 5 and d = 2.

∴ nth term = a + (n – 1) d = 5 + (n – 1) 2 = (2n + 3)

∴ nth term, t_{n} = \(\frac{1}{(2 n+1)(2 n+3)}\)

P(n) ≡ \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}\) = \(\frac{n}{3(2 n+3)}\)

Step I:

Put n = 1

L.H.S. = \(\frac{1}{3.5}=\frac{1}{15}\)

R.H.S. = \(\frac{1}{3[2(1)+3]}=\frac{1}{3(2+3)}=\frac{1}{15}\)

∴ 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.

∴ \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 k+1)(2 k+3)}\) = \(\frac{k}{3(2 k+3)}\) ….(i)

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.

∴ \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots \text { to } n \text { terms }=\frac{n}{3(2 n+3)}\), for all n ∈ N.

Question 10.

(2^{3n} – 1) is divisible by 7.

Solution:

(2^{3n} – 1) is divisible by 7 if and only if (2^{3n} – 1) is a multiple of 7.

Let P(n) ≡ (2^{3n} – 1) = 7m, where m ∈ N.

Step I:

Put n = 1

∴ 2^{3n} – 1 = 2^{3(1)} – 1 = 2^{3} – 1 = 8 – 1 = 7

∴ (2^{3n} – 1) is a multiple of 7.

∴ P(n) is true for n = 1.

Step II:

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

i.e., 2^{3k} – 1 is a multiple of 7.

∴ 2^{3k} – 1 = 7a, where a ∈ N

∴ 2^{3k} = 7a + 1 ……(i)

Step III:

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

i.e., to prove that

2^{3(k+1)} – 1 = 7b, where b ∈ N.

∴ P(k + 1) = 2^{3(k+1)} – 1

= 2^{3k+3 }– 1

= 2^{3k} . (2^{3}) – 1

= (7a + 1)8 – 1 …..[From (i)]

= 56a + 8 – 1

= 56a + 7

= 7(8a + 1)

7b, where b = (8a + 1) ∈ N

∴ 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^{4n} – 1) is divisible by 7, for all n ∈ N.

Question 11.

(2^{4n} – 1) is divisible by 15.

Solution:

(2^{4n} – 1) is divisible by 15 if and only if (2^{4n} – 1) is a multiple of 15.

Let P(n) ≡ (2^{4n} – 1) = 15m, where m ∈ N.

Step I:

Put n = 1

∴ 2^{4(1)} – 1 = 16 – 1 = 15

∴ (2^{4n} – 1) is a multiple of 15.

∴ P(n) is true for n = 1.

Step II:

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

∴ 2^{4k} – 1 = 15a, where a ∈ N

∴ 2^{4k} = 15a + 1 …..(i)

Step III:

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

i.e., to prove that

∴ 2^{4(k+1)} – 1 = 15b, where b ∈ N

∴ P(k + 1) = 2^{4(k+1)} – 1 = 2^{4k+4} – 1

= 2^{4k} . 2^{4} – 1

= 16 . (2^{4k}) – 1

= 16(15a + 1) – 1 …..[From (i)]

= 240a + 16 – 1

= 240a + 15

= 15(16a + 1)

= 15b, where b = (16a + 1) ∈ N

∴ 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^{4n} – 1) is divisible by 15, for all n ∈ N.

Question 12.

3^{n} – 2n – 1 is divisible by 4.

Solution:

(3^{n} – 2n – 1) is divisible by 4 if and only if (3^{n} – 2n – 1) is a multiple of 4.

Let P(n) ≡ (3^{n} – 2n – 1) = 4m, where m ∈ N.

Step I:

Put n = 1

∴ (3^{n} – 2n – 1) = 3^{(1)} – 2(1) – 1 = 0 = 4(0)

∴ (3^{n} – 2n – 1) is a multiple of 4.

∴ P(n) is tme for n = 1.

Step II:

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

∴ 3^{k} – 2k – 1 = 4a, where a ∈ N

∴ 3^{k} = 4a + 2k + 1 ….(i)

Step III:

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

i.e., to prove that

3^{(k+1)} – 2(k + 1) – 1 = 4b, where b ∈ N

P(k + 1) = 3^{k+1} – 2(k + 1) – 1

= 3^{k} . 3 – 2k – 2 – 1

= (4a + 2k + 1) . 3 – 2k – 3 …….[From (i)]

= 12a + 6k + 3 – 2k – 3

= 12a + 4k

= 4(3a + k)

= 4b, where b = (3a + k) ∈ N

∴ P(n) is tme for n = k + 1.

Step IV:

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

∴ 3^{n} – 2n – 1 is divisible by 4, for all n ∈ N.

Question 13.

5 + 5^{2} + 5^{3} + ….. + 5^{n} = \(\frac{5}{4}\) (5^{n} – 1)

Solution:

Let P(n) ≡ 5 + 52 + 53 +…..+ 5n = \(\frac{5}{4}\) (5n – 1), for all n ∈ N.

Step I:

Put n = 1

L.H.S. = 5

R.H.S. = \(\frac{5}{4}\) (5^{1} – 1) = 5

∴ 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.

∴ 5 + 5^{2} + 5^{3} + ….. + 5^{k} = \(\frac{5}{4}\) (5^{k} – 1) …….(i)

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.

∴ 5 + 5^{2} + 5^{3} + … + 5^{n} = \(\frac{5}{4}\) (5^{n} – 1), for all n ∈ N.

Question 14.

(cos θ + i sin θ)^{n} = cos (nθ) + i sin (nθ)

Solution:

Let P(n) ≡ (cos θ + i sin θ)^{n} = cos nθ + i sin nθ, for all n ∈ N.

Step I:

Put n = 1

L.H.S. = (cos θ + i sin θ)^{1} = cos θ + i sin θ

R.H.S. = cos[(1)θ] + i sin[(1)θ] = cos θ + i sin θ

∴ 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.

∴ (cos θ + i sin θ)^{k} = cos kθ + i sin kθ …….(i)

Step III:

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

i.e., to prove that

(cos θ + i sin θ)^{k+1} = cos (k + 1)θ + i sin (k + 1)θ

L.H.S. = (cos θ + i sin θ)^{k+1}

= (cos θ + i sin θ)^{k} . (cos θ + i sin θ)

= (cos kθ + i sin kθ) . (cos θ + i sin θ) ……[From (i)]

= cos kθ cos θ + i sin θ cos kθ + i sin kθ cosθ – sin kθ sin θ ……[∵ i^{2} = -1]

= (cos kθ cos θ – sin k θ sin θ) + i(sin kθ cos θ + cos kθ sin θ)

= cos(kθ + θ) + i sin(kθ + θ)

= cos(k + 1) θ + i sin (k + 1) θ

= R.H.S.

∴ 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.

∴ (cos θ + i sin θ)^{n} = cos (nθ) + i sin (nθ), for all n ∈ N.

Question 15.

Given that t_{n+1} = 5 t_{n+4}, t_{1} = 4, prove by method of induction that t_{n} = 5^{n} – 1.

Solution:

Let the statement P(n) has L.H.S. a recurrence relation t_{n+1} = 5 t_{n+4}, t_{1} = 4 and R.H.S. a general statement t_{n} = 5^{n} – 1.

Step I:

Put n = 1

L.H.S. = 4

R.H.S. = 5^{1} – 1 = 4

∴ L.H.S. = R.H.S.

∴ P(n) is true for n = 1.

Put n = 2

L.H.S. = t_{2} = 5t_{1} + 4 = 24

R.H.S. = t_{2} = 5^{2} – 1 = 24

∴ L.H.S. = R.H.S.

∴ P(n) is true for n = 2.

Step II:

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

∴ t_{k+1} = 5 t_{k+4} and t_{k} = 5^{k} – 1

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

Since t_{k+1} = 5 t_{k+4} and t_{k} = 5^{k} – 1 …..[From Step II]

t_{k+1} = 5 (5^{k} – 1) + 4 = 5^{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.

∴ t_{n} = 5^{n} – 1, for all n ∈ N.

Question 16.

Prove by method of induction

\(\left(\begin{array}{ll}

1 & 2 \\

0 & 1

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

1 & 2 n \\

0 & 1

\end{array}\right) \forall n \in N\)

Solution: