**Chapter 4 Methods of Induction and Binomial Theorem Ex 4.1**

## 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.**

Step II:

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

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.

**Question 4.**

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.

Step II:

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

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.

**Question 6.**

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:

Step 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 = (4n^{2} + 6n – 1) for all n ∈ N.

**Question 8.**

Step II:

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

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.

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

Step I:

Step II:

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

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.

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

**Question 10.**

Step III:

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.

Step III:

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.**

Step III:

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

i.e., to prove that

= (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.**

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.

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

**Question 14.**

Step III:

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

i.e., to prove that

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)^{2} = cos (nθ) + i sin (nθ), for all n ∈ N.

**Question 15.**

Step III:

We have 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.

∴ t_{n} =5^{n }– 1, for all n ∈ N.

**Question 16.Prove by method of inductionSolution:**