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 = n(n+1)(4n1)3

Solution:

From the statement formula

When n = 1,

LHS =1 x 2 = 2

RHS = 1(1+1)(4x11)3 = 63 = 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 = k(k+1)(4k1)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)

= k(k+1)(4k1)3 + (2(k + 1) - 1) x 2(k + 1)

= (k+1)3(4k2 - k + 12 k + 6)

= (k+1)(4k2+8k+3k+6)3

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

= (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 

From Problems on Principle of Mathematical Induction to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.




Share this page: What’s this?

Recent Articles

  1. 5th Grade Circle Worksheet | Free Worksheet with Answer |Practice Math

    Jul 11, 25 02:14 PM

    Radii of the circRadii, Chords, Diameters, Semi-circles
    In 5th Grade Circle Worksheet you will get different types of questions on parts of a circle, relation between radius and diameter, interior of a circle, exterior of a circle and construction of circl…

    Read More

  2. Construction of a Circle | Working Rules | Step-by-step Explanation |

    Jul 09, 25 01:29 AM

    Parts of a Circle
    Construction of a Circle when the length of its Radius is given. Working Rules | Step I: Open the compass such that its pointer be put on initial point (i.e. O) of ruler / scale and the pencil-end be…

    Read More

  3. Combination of Addition and Subtraction | Mixed Addition & Subtraction

    Jul 08, 25 02:32 PM

    Add and Sub
    We will discuss here about the combination of addition and subtraction. The rules which can be used to solve the sums involving addition (+) and subtraction (-) together are: I: First add

    Read More

  4. Addition & Subtraction Together |Combination of addition & subtraction

    Jul 08, 25 02:23 PM

    Addition and Subtraction Together Problem
    We will solve the different types of problems involving addition and subtraction together. To show the problem involving both addition and subtraction, we first group all the numbers with ‘+’ and…

    Read More

  5. 5th Grade Circle | Radius, Interior and Exterior of a Circle|Worksheet

    Jul 08, 25 09:55 AM

    Semi-circular Region
    A circle is the set of all those point in a plane whose distance from a fixed point remains constant. The fixed point is called the centre of the circle and the constant distance is known

    Read More