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





11 and 12 Grade Math 

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