7 Most FAQ’s of Mathematical Induction Chapter in Inter 1st Year Maths-1A (TS/AP)

7 Marks

LAQ-1 : Using the principle of finite Mathematical Induction prove that 1.2.3+2.3.4+3.4.5.+…. Upto n terms = (n(n+1)(n+2)(n+3))/4, ∀ n ∈ N.

Given Series: The nth term of the given series is $$T_n = n(n + 1)(n + 2)$$

To Find: We need to show that the sum of the series up to the nth term, denoted as S(n), is $$S(n) = \frac{n(n+1)(n+2)(n+3)}{4}$$

Proof:

Step 1: Base Case

Let’s start with the base case, S(1):

$$L.H.S = 1 \cdot 2 \cdot 3 = 6$$

$$R.H.S = \frac{1(1+1)(1+2)(1+3)}{4} = \frac{2 \cdot 3 \cdot 4}{4} = 6$$

Since $$L.H.S = R.H.S$$ the statement holds true for $$n = 1$$

Step 2: Inductive Hypothesis

Assume that the statement is true for some arbitrary natural number k, i.e.,

$$S(k) = 1\cdot2\cdot3 + 2\cdot3\cdot4 + \ldots + k(k+1)(k+2) = \frac{k(k+1)(k+2)(k+3)}{4} \quad \dots (1)$$

Step 3: Inductive Step

We need to prove that the statement holds for k+1, i.e., S(k+1) is true.

Adding $$(k+1)(k+2)(k+3)$$ to both sides of equation (1), we get:

$$L.H.S = [1\cdot2\cdot3 + 2\cdot3\cdot4 + \ldots + k(k+1)(k+2)] + (k+1)(k+2)(k+3)$$

$$= \frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3)$$

$$= \frac{k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3)}{4}$$

$$= \frac{(k+1)(k+2)(k+3)(k+4)}{4} = R.H.S$$

Since $$L.H.S = R.H.S$$ our assumption that the statement holds true for k implies that it also holds true for k+1.

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LAQ-2 : Using the principle of finite Mathematical Induction prove that 1²+(1²+2² )+(1²+2²+3² )+…n terms=  (n(n+1)² (n+2))/12, ∀ n ∈ N.

Given Series: The nth term is the sum of the squares of the first n numbers, expressed as $$T_n = 1^2 + 2^2 + 3^2 + \ldots + n^2$$

To Find: We aim to prove that the sum of this series up to the nth term, denoted as S(n), is $$S(n) = \frac{n(n+1)(2n+1)}{6}$$ and subsequently that $$S(n) = \frac{n(n+1)^2(n+2)}{12}$$

Proof:

Step 1: Base Case

For $$n = 1$$

$$L.H.S = 1^2 = 1$$

$$R.H.S = \frac{1(1+1)^2(1+2)}{12} = \frac{1 \cdot 2^2 \cdot 3}{12} = \frac{12}{12} = 1$$

Since $$L.H.S = R.H.S$$ the base case S(1) is true.

Step 2: Inductive Hypothesis

Assume that the statement is true for some positive integer k, i.e.,

$$S(k) = 1^2 + (1^2 + 2^2) + \ldots + \frac{k(k+1)(2k+1)}{6} = \frac{k(k+1)^2(k+2)}{12} \quad \dots (1)$$

Step 3: Inductive Step

We need to prove that the statement holds for k+1, i.e., S(k+1) is true.

The (k+1)th term is $$\frac{(k+1)(k+2)(2(k+1)+1)}{6} = \frac{(k+1)(k+2)(2k+3)}{6}$$

Adding the (k+1)th term to both sides of equation (1), we get:

$$L.H.S = [1^2 + (1^2 + 2^2) + \ldots + \frac{k(k+1)(2k+1)}{6}] + \frac{(k+1)(k+2)(2k+3)}{6}​$$

$$= \frac{k(k+1)^2(k+2)}{12} + \frac{(k+1)(k+2)(2k+3)}{6}$$

$$= \frac{k(k+1)^2(k+2)+2(k+1)(k+2)(2k+3)}{12}$$

$$= \frac{(k+1)(k+2)[k(k+1)+2(2k+3)]}{12}$$

$$= \frac{(k+1)(k+2)(k^2+5k+6)}{12}$$

$$= \frac{(k+1)(k+2)^2(k+3)}{12} = R.H.S$$

Since $$L.H.S = R.H.S$$ our assumption that the statement holds true for k implies that it also holds true for k+1.

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LAQ-3 : Show that 1/1.3+1/3.5+1/5.7+…+n terms = n/(2n+1)

Given Series: The series is formed by the reciprocals of the product of consecutive odd numbers, starting from 1. The nth term is represented as $$T_n = \frac{1}{(2n – 1)(2n + 1)}$$

To Find: We aim to prove that the sum of the series up to the nth term, denoted as S(n), is $$S(n) = \frac{n}{2n + 1}$$

Proof:

Step 1: Base Case

For $$n = 1$$

$$L.H.S = \frac{1}{1 \cdot 3} = \frac{1}{3}$$

$$R.H.S = \frac{1}{2 \cdot 1 + 1} = \frac{1}{3}$$

Since $$L.H.S = R.H.S$$ the base case S(1) is true.

Step 2: Inductive Hypothesis

Assume that the statement is true for some positive integer k, i.e.,

$$S(k) = \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \ldots + \frac{1}{(2k – 1)(2k + 1)} = \frac{k}{2k + 1} \quad \dots (1)$$

Step 3: Inductive Step

We need to prove that the statement holds for k+1, i.e., S(k+1) is true.

The (k+1)th term is $$\frac{1}{(2(k+1) – 1)(2(k+1) + 1)} = \frac{1}{(2k + 1)(2k + 3)}$$

Adding the (k+1)th term to both sides of equation (1), we get:

$$L.H.S = \left[\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \ldots + \frac{1}{(2k – 1)(2k + 1)}\right] + \frac{1}{(2k + 1)(2k + 3)}$$

$$= \frac{k}{2k + 1} + \frac{1}{(2k + 1)(2k + 3)}$$

To simplify, notice that:

$$= \frac{k(2k + 3) + 1}{(2k + 1)(2k + 3)}$$

$$= \frac{2k^2 + 3k + 1}{(2k + 1)(2k + 3)}$$

$$= \frac{(k + 1)(2k + 1)}{(2k + 1)(2k + 3)}$$

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

Since $$L.H.S = R.H.S$$ our assumption that the statement holds true for k implies that it also holds true for k+1.

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LAQ-4 : Show that 1/1.4 + 1/4.7 + 1/7.10 +… + n terms =  n/(3n+1).

Given Series: The series is formed by the reciprocals of the product of terms from two related APs, with the nth term represented as $$T_n = \frac{1}{(3n – 2)(3n + 1)}$$

To Find: We aim to prove that the sum of the series up to the nth term, denoted as S(n), equals $$S(n) = \frac{n}{3n + 1}$$

Proof:

Step 1: Base Case

For $$n = 1$$

$$L.H.S = \frac{1}{1 \cdot 4} = \frac{1}{4}$$

$$R.H.S = \frac{1}{3 \cdot 1 + 1} = \frac{1}{4}$$

Since $$L.H.S = R.H.S$$ the base case S(1) is true.

Step 2: Inductive Hypothesis

Assume the statement is true for some positive integer k, i.e.,

$$S(k) = \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \ldots + \frac{1}{(3k – 2)(3k + 1)} = \frac{k}{3k + 1} \quad \dots (1)$$

Step 3: Inductive Step

We need to prove the statement holds for k+1, i.e., S(k+1) is true.

The (k+1)th term is $$\frac{1}{(3(k+1) – 2)(3(k+1) + 1)} = \frac{1}{(3k + 1)(3k + 4)}$$

Adding the (k+1)th term to both sides of equation (1), we get:

$$L.H.S = \left[\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \ldots + \frac{1}{(3k – 2)(3k + 1)}\right] + \frac{1}{(3k + 1)(3k + 4)}$$

$$= \frac{k}{3k + 1} + \frac{1}{(3k + 1)(3k + 4)}$$

$$= \frac{k(3k + 4) + 1}{(3k + 1)(3k + 4)}$$

$$= \frac{3k^2 + 4k + 1}{(3k + 1)(3k + 4)}$$

$$= \frac{(k + 1)(3k + 1)}{(3k + 1)(3k + 4)}$$

$$= \frac{k + 1}{3k + 4} = \frac{k + 1}{3(k + 1) + 1} = R.H.S$$

Since $$L.H.S = R.H.S$$ our assumption that the statement holds true for k implies it also holds true for k+1.

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LAQ-5 : Show that 1³/1 +(1³ + 2³)/(1 + 3)+…n terms=  n/24(2n² + 9n + 13).

Given Series: The series involves the ratio of the sum of cubes to the sum of the first n odd numbers. The nth term is derived as $$T_n = \frac{n^2(n + 1)^2}{4n^2} = \frac{(n + 1)^2}{4}$$

To Find: We aim to prove that the sum of the series up to the nth term, denoted as S(n), equals $$S(n) = \frac{n}{24}(2n^2 + 9n + 13)$$

Proof:

Step 1: Base Case

For $$n = 1$$

$$L.H.S = \frac{1^3}{1} = 1$$

$$R.H.S = \frac{1}{24}(2 \cdot 1^2 + 9 \cdot 1 + 13) = \frac{24}{24} = 1$$

Since $$L.H.S = R.H.S$$ the base case S(1) is true.

Step 2: Inductive Hypothesis

Assume the statement is true for some positive integer k, i.e.,

$$S(k) = \frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \ldots + \frac{(k + 1)^2}{4} = \frac{k}{24}(2k^2 + 9k + 13) \quad \dots (1)$$

Step 3: Inductive Step

We need to prove the statement holds for k+1, i.e., S(k+1) is true.

Adding the (k+1)^th term to both sides of equation (1), we get:

$$L.H.S = \left[ \frac{(k + 1)^2}{4} \right] + \frac{(k + 2)^2}{4}$$

$$= \frac{k}{24}(2k^2 + 9k + 13) + \frac{(k + 2)^2}{4}$$

$$= \frac{k(2k^2 + 9k + 13) + 6(k^2 + 4k + 4)}{24}$$

$$= \frac{2k^3 + 9k^2 + 13k + 6k^2 + 24k + 24}{24}$$

$$= \frac{2k^3 + 15k^2 + 37k + 24}{24} \quad \dots (2)$$

To derive R.H.S for S(k+1):

$$R.H.S = \frac{(k + 1)}{24}(2(k + 1)^2 + 9(k + 1) + 13)$$

$$= \frac{(k + 1)}{24}(2k^2 + 4k + 2 + 9k + 9 + 13)$$

$$= \frac{(k + 1)}{24}(2k^2 + 13k + 24)$$

$$= \frac{2k^3 + 13k^2 + 24k + 2k^2 + 13k + 24}{24}$$

$$= \frac{2k^3 + 15k^2 + 37k + 24}{24} \quad \dots (3)$$

From equations 22 and 33, we see $$L.H.S = R.H.S$$

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LAQ-6 : By Mathematical Induction, show that 49ⁿ + 16n – 1 is divisible by 64 for all positive integer n.

Given Statement: $$S(n) : 49^n + 16^n – 1 = 64q$$ where $$q \in \mathbb{Z}$$ (the set of all integers).

Proof:

Step 1: Base Case

For $$n = 1$$

$$L.H.S = 49(1) + 16(1) – 1 = 49 + 16 – 1 = 64$$

$$= 64 \times 1$$

Since 64 is divisible by 64, S(1) is true.

Step 2: Inductive Hypothesis

Assume S(k) is true for some positive integer k, i.e.,

$$S(k) : 49^k + 16^k – 1 = 64q \quad \dots (1)$$

where q is an integer.

Step 3: Inductive Step

To prove S(k+1) is true, consider the expression for k+1:

$$L.H.S = 49^{k+1} + 16^{k+1} – 1$$

Using the expression from S(k), we expand L.H.S as follows:

$$= 49^k \cdot 49 + 16^k \cdot 16 – 1$$

Using the assumption from S(k) that $$49^k + 16^k – 1 = 64q$$

$$= (64q + 16^k – 1) \cdot 49 + 16^k \cdot 16 – 1$$

$$= 64q \cdot 49 + 49 \cdot 16^k – 49 + 16^k \cdot 16 – 1$$

Rearrange and factor out the common terms:

$$= 64q \cdot 49 + 16^k(49 + 16) – 50$$

$$= 64q \cdot 49 + 16^k \cdot 65 – 50$$

Observe that 64q⋅49 is divisible by 64, 16^k⋅65 can be written as $$16^k \cdot 64 + 16^k$$ making it also divisible by 64, and adjust the term −50 to fit the divisibility:

$$= 64q \cdot 49 + 16^k \cdot 64 + 16^k – 50$$

$$= 64(q \cdot 49 + 16^k) + (16^k – 50)$$

Thus, we demonstrate that if S(k) is true, then S(k+1) is also true.

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LAQ-7 : Using the principle of Mathematical Induction, prove that 3.5⁽²ⁿ⁺¹⁾+2⁽³ⁿ⁺¹⁾  is divisible by 17, ∀ n ∈ N.

Given Statement: $$S(n) : 3 \cdot 5^{2n+1} + 2^{3n+1} = 17q$$ where $$q \in \mathbb{Z}$$ (the set of all integers).

Proof:

Step 1: Base Case

For $$n = 1$$

$$L.H.S = 3 \cdot 5^{2(1)+1} + 2^{3(1)+1} = 3 \cdot 5^3 + 2^4 = 3 \cdot 125 + 16 = 375 + 16 = 391$$

$$391 = 17 \cdot 23$$

Since 391 is divisible by 17, S(1) is true.

Step 2: Inductive Hypothesis

Assume S(k) is true for some positive integer k, i.e.,

$$S(k) : 3 \cdot 5^{2k+1} + 2^{3k+1} = 17q \quad \dots (1)$$

where q is an integer.

Step 3: Inductive Step

To prove S(k+1) is true, consider the expression for k+1:

$$L.H.S = 3 \cdot 5^{2(k+1)+1} + 2^{3(k+1)+1}$$

$$= 3 \cdot 5^{2k+2+1} + 2^{3k+3+1}$$

$$= 3 \cdot 5^{2k+1} \cdot 5^2 + 2^{3k+1} \cdot 2^3$$

Expanding the expression using the given S(k) and the values for 5² and 2³:

$$= 3 \cdot 5^{2k+1} \cdot 25 + 2^{3k+1} \cdot 8$$

Utilizing the assumption from S(k) that

$$3 \cdot 5^{2k+1} + 2^{3k+1} = 17q$$

$$= 25 \cdot (3 \cdot 5^{2k+1} + 2^{3k+1}) – 3 \cdot 5^{2k+1} \cdot 17 + 2^{3k+1} \cdot 8$$

$$= 25 \cdot 17q – 17 \cdot (3 \cdot 5^{2k+1} – 2^{3k+1})$$

S(k+1) is true whenever S(k) is true.