# Problems on Principle of Mathematical Induction

Solved Problems on Principle of Mathematical Induction are shown here to prove Mathematical Induction.

### Problems on Principle of Mathematical Induction

1. Using the principle of mathematical induction, prove that

1² + 2² + 3² + ..... + n² = (1/6){n(n + 1)(2n + 1} for all n ∈ N.

Solution:

Let the given statement be P(n). Then,

P(n): 1² + 2² + 3² + ..... +n² = (1/6){n(n + 1)(2n + 1)}.

Putting n =1 in the given statement, we get

LHS = 1² = 1 and RHS = (1/6) × 1 × 2 × (2 × 1 + 1) = 1.

Therefore LHS = RHS.

Thus, P(1) is true.

Let P(k) be true. Then,

P(k): 1² + 2² + 3² + ..... + k² = (1/6){k(k + 1)(2k + 1)}.

Now, 1² + 2² + 3² + ......... + k² + (k + 1)²

= (1/6) {k(k + 1)(2k + 1) + (k + 1)²

= (1/6){(k + 1).(k(2k + 1)+6(k + 1))}

= (1/6){(k + 1)(2k² + 7k + 6})

= (1/6){(k + 1)(k + 2)(2k + 3)}

= 1/6{(k + 1)(k + 1 + 1)[2(k + 1) + 1]}

⇒ P(k + 1): 1² + 2² + 3² + ….. + k² + (k+1)²

= (1/6){(k + 1)(k + 1 + 1)[2(k + 1) + 1]}

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

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

2. By using mathematical induction prove that the given equation is true for all positive integers.

1 x 2 + 3 x 4 + 5 x 6 + …. + (2n - 1) x 2n = $$\frac{n (n + 1) (4n - 1)}{3}$$

Solution:

From the statement formula

When n = 1,

LHS =1 x 2 = 2

RHS = $$\frac{1(1 + 1) (4 x 1 - 1)}{3}$$ = $$\frac{6}{3}$$ = 2

Hence it is proved that P (1) is true for the equation.

Now we assume that P (k) is true or 1 x 2 + 3 x 4 + 5 x 6 + …. + (2k - 1) x 2k = $$\frac{k(k + 1)(4k - 1)}{3}$$.

For P(k + 1)

LHS = 1 x 2 + 3 x 4 + 5 x 6 + …. + (2k - 1) x 2k + (2(k + 1) - 1) x 2(k + 1)

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

= $$\frac{(k + 1)}{3}$$(4k2 - k + 12 k + 6)

= $$\frac{(k + 1) (4k^{2} + 8k + 3k + 6)}{3}$$

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

= $$\frac{(k + 1)((k + 1) + 1)(4(k + 1) - 1)}{3}$$ = RHS for P (k+1)

Now it is proved that P (k + 1) is also true for the equation.

So the given statement is true for all positive integers.

Problems on Principle of Mathematical Induction

3. Using the principle of mathematical induction, prove that

1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ..... + n(n + 1) = (1/3){n(n + 1)(n + 2)}.

Solution:

Let the given statement be P(n). Then,

P(n): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ..... + n(n + 1) = (1/3){n(n + 1)(n + 2)}.

Thus, the given statement is true for n = 1, i.e., P(1) is true.

Let P(k) be true. Then,

P(k): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ..... + k(k + 1) = (1/3){k(k + 1)(k + 2)}.

Now, 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +...+ k(k + 1) + (k + 1)(k + 2)

= (1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ....... + k(k + 1)) + (k + 1)(k + 2)

= (1/3) k(k + 1)(k + 2) + (k + 1)(k + 2) [using (i)]

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

= (1/3){(k + 1)(k + 2)(k + 3)}

⇒ P(k + 1): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +......+ (k + 1)(k + 2)

= (1/3){k + 1 )(k + 2)(k +3)}

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1)is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all values of ∈ N.

Problems on Principle of Mathematical Induction

4. By using mathematical induction prove that the given equation is true for all positive integers.

2 + 4 + 6 + …. + 2n = n(n+1)

Solution:

From the statement formula

When n = 1 or P (1),

LHS = 2

RHS =1 × 2 = 2

So P (1) is true.

Now we assume that P (k) is true or 2 + 4 + 6 + …. + 2k = k(k + 1).

For P(k + 1),

LHS = 2 + 4 + 6 + …. + 2k + 2(k + 1)

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

= (k + 1) (k + 2)

= (k + 1) ((k + 1) + 1) = RHS for P(k+1)

Now it is proved that P(k+1) is also true for the equation.

So the given statement is true for all positive integers.

5. Using the principle of mathematical induction, prove that

1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 +.....+ (2n - 1)(2n + 1) = (1/3){n(4n² + 6n - 1).

Solution:

Let the given statement be P(n). Then,

P(n): 1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 +...... + (2n - 1)(2n + 1)= (1/3)n(4n² + 6n - 1).

When n = 1, LHS = 1 ∙ 3 = 3 and RHS = (1/3) × 1 × (4 × 1² + 6 × 1 - 1)

= {(1/3) × 1 × 9} = 3.

LHS = RHS.

Thus, P(1) is true.

Let P(k) be true. Then,

P(k): 1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 + ….. + (2k - 1)(2k + 1) = (1/3){k(4k² + 6k - 1) ......(i)

Now,

1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 + …….. + (2k - 1)(2k + 1) + {2k(k + 1) - 1}{2(k + 1) + 1}

= {1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 + ………… + (2k - 1)(2k + 1)} + (2k + 1)(2k + 3)

= (1/3) k(4k² + 6k - 1) + (2k + 1)(2k + 3) [using (i)]

= (1/3) [(4k³ + 6k² - k) + 3(4k² + 8k + 3)]

= (1/3)(4k³ + 18k² + 23k + 9)

= (1/3){(k + 1)(4k² + 14k + 9)}

= (1/3)[k + 1){4k(k + 1) ² + 6(k + 1) - 1}]

⇒ P(k + 1): 1 ∙ 3 + 3 ∙ 5 + 5 ∙ 7 + ..... + (2k + 1)(2k + 3)

= (1/3)[(k + 1){4(k + 1)² + 6(k + 1) - 1)}]

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

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

More Problems on Principle of Mathematical Induction

6. By using mathematical induction prove that the given equation is true for all positive integers.

2 + 6 + 10 + ….. + (4n - 2) = 2n2

Solution:

From the statement formula

When n = 1 or P(1),

LHS = 2

RHS = 2 × 12 = 2

So P(1) is true.

Now we assume that P (k) is true or 2 + 6 + 10 + ….. + (4k - 2) = 2k2

For P (k + 1),

LHS = 2 + 6 + 10 + ….. + (4k - 2) + (4(k + 1) - 2)

= 2k2 + (4k + 4 - 2)

= 2k+ 4k + 2

= (k+1)2

= RHS for P(k+1)

Now it is proved that P(k+1) is also true for the equation.

So the given statement is true for all positive integers.

7. Using the principle of mathematical induction, prove that

1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ..... + 1/{n(n + 1)} = n/(n + 1)

Solution:

Let the given statement be P(n). Then,

P(n): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ..... + 1/{n(n + 1)} = n/(n + 1).

Putting n = 1 in the given statement, we get

LHS= 1/(1 ∙ 2) = and RHS = 1/(1 + 1) = 1/2.

LHS = RHS.

Thus, P(1) is true.

Let P(k) be true. Then,

P(k): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ..... + 1/{k(k + 1)} = k/(k + 1) ..…(i)

Now 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ..... + 1/{k(k + 1)} + 1/{(k + 1)(k + 2)}

[1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ..... + 1/{k(k + 1)}] + 1/{(k + 1)(k + 2)}

= k/(k + 1)+1/{ (k + 1)(k + 2)}.

{k(k + 2) + 1}/{(k + 1)²/[(k + 1)k + 2)] using …(ii)

= {k(k + 2) + 1}/{(k + 1)(k + 2}

= {(k + 1)² }/{(k + 1)(k + 2)}

= (k + 1)/(k + 2) = (k + 1)/(k + 1 + 1)

⇒ P(k + 1): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ……… + 1/{ k(k + 1)} + 1/{(k + 1)(k + 2)}

= (k + 1)/(k + 1 + 1)

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1)is true, whenever P(k) is true.

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

Problems on Principle of Mathematical Induction

8. Using the principle of mathematical induction, prove that

{1/(3 ∙ 5)} + {1/(5 ∙ 7)} + {1/(7 ∙ 9)} + ….... + 1/{(2n + 1)(2n + 3)} = n/{3(2n + 3)}.

Solution:

Let the given statement be P(n). Then,

P(n): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + ……. + 1/{(2n + 1)(2n + 3)} = n/{3(2n + 3).

Putting n = 1 in the given statement, we get

and LHS = 1/(3 ∙ 5) = 1/15 and RHS = 1/{3(2 × 1 + 3)} = 1/15.

LHS = RHS

Thus , P(1) is true.

Let P(k) be true. Then,

P(k): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + …….. + 1/{(2k + 1)(2k + 3)} = k/{3(2k + 3)} ….. (i)

Now, 1/(3 ∙ 5) + 1/(5 ∙ 7) + ..…… + 1/[(2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}2(k + 1) + 3

= {1/(3 ∙ 5) + 1/(5 ∙ 7) + ……. + [1/(2k + 1)(2k + 3)]} + 1/{(2k + 3)(2k + 5)}

= k/[3(2k + 3)] + 1/[2k + 3)(2k + 5)] [using (i)]

= {k(2k + 5) + 3}/{3(2k + 3)(2k + 5)}

= (2k² + 5k + 3)/[3(2k + 3)(2k + 5)]

= {(k + 1)(2k + 3)}/{3(2k + 3)(2k + 5)}

= (k + 1)/{3(2k + 5)}

= (k + 1)/[3{2(k + 1) + 3}]

= P(k + 1): 1/(3 ∙ 5) + 1/(5 ∙ 7) + …….. + 1/[2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}{2(k + 1) + 3}]

= (k + 1)/{3{2(k + 1) + 3}]

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for n ∈ N.

Problems on Principle of Mathematical Induction
9. By induction prove that 3- 1 is divisible by 2 is true for all positive integers.

Solution:

When n = 1, P(1) = 31 - 1 = 2 which is divisible by 2.

So P(1) is true.

Now we assume that P(k) is true or 3k - 1 is divisible by 2.

When P(k + 1),

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

As (3k - 1) and 2 both are divisible by 2, it is proved that divisible by 2 is true for all positive integers.

10. Using the principle of mathematical induction, prove that

1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + …….. + 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)} for all n ∈ N.

Solution:

Let P (n): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……. + 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)} .

Putting n = 1 in the given statement, we get

LHS = 1/(1 ∙ 2 ∙ 3) = 1/6 and RHS = {1 × (1 + 3)}/[4 × (1 + 1)(1 + 2)] = ( 1 × 4)/(4 × 2 × 3) = 1/6.

Therefore LHS = RHS.

Thus, the given statement is true for n = 1, i.e., P(1) is true.

Let P(k) be true. Then,

P(k): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……... + 1/{k(k + 1)(k + 2)} = {k(k + 3)}/{4(k + 1)(k + 2)}. …….(i)

Now, 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………….. + 1/{k(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}

= [1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………..…. + 1/{ k(k + 1)(k + 2}] + 1/{(k + 1)(k + 2)(k + 3)}

= [{k(k + 3)}/{4(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}]
[using(i)]

= {k(k + 3)² + 4}/{4(k + 1)(k + 2)(k + 3)}

= (k³ + 6k² + 9k + 4)/{4(k + 1)(k + 2)(k + 3)}

= {(k + 1)(k + 1)(k + 4)}/{4 (k + 1)(k + 2)(k + 3)}

= {(k + 1)(k + 4)}/{4(k + 2)(k + 3)

⇒ P(k + 1): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……….….. + 1/{(k + 1)(k + 2)(k + 3)}

= {(k + 1)(k + 2)}/{4(k + 2)(k + 3)}

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

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

Problems on Principle of Mathematical Induction

11. By induction prove that n- 3n + 4 is even and it is true for all positive integers.

Solution:

When n = 1, P (1) = 1 - 3 + 4 = 2 which is an even number.

So P (1) is true.

Now we assume that P (k) is true or k- 3k + 4 is an even number.

When P (k + 1),

(k + 1)- 3(k + 1) + 4

= k+ 2k + 1 - 3k + 3 + 4

= k- 3k + 4 + 2(k + 2)

As k- 3k + 4 and 2(k + 2) both are even, there sum also will be an even number.

So it is proved that n- 3n + 4 is even is true for all positive integers.

12. Using the Principle of mathematical induction, prove that

{1 - (1/2)}{1 - (1/3)}{1 - (1/4)} ….... {1 - 1/(n + 1)} = 1/(n + 1) for all n ∈ N.

Solution:

Let the given statement be P(n). Then,

P(n): {1 - (1/2)}{1 - (1/3)}{1 - (1/4)} ….... {1 - 1/(n + 1)} = 1/(n + 1).

When n = 1, LHS = {1 – (1/2)} = ½ and RHS = 1/(1 + 1) = ½.

Therefore LHS = RHS.

Thus, P(1) is true.

Let P(k) be true. Then,

P(k): {1 - (1/2)}{1 - (1/3)}{1 - (1/4)} ….... [1 - {1/(k + 1)}] = 1/(k + 1)

Now, [{1 - (1/2)}{1 - (1/3)}{1 - (1/4)} ….... [1 - {1/(k + 1)}] ∙ [1 – {1/(k + 2)}]

= [1/(k + 1)] ∙ [{(k + 2 ) - 1}/(k + 2)}]

= [1/(k + 1)] ∙ [(k + 1)/(k + 2)]

= 1/(k + 2)

Therefore p(k + 1): [{1 - (1/2)}{1 - (1/3)}{1 - (1/4)} ….... [1 - {1/(k + 1)}] = 1/(k + 2)

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

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

Problems on Principle of Mathematical Induction

Mathematical Induction