MATHEMATICAL INDUCTION PROBLEMS WITH SOLUTIONS

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Mathematical Induction Problems With Solutions :

Here we are going to see some mathematical induction problems with solutions.

Define mathematical induction :

Mathematical Induction is a method or technique of proving mathematical results or theorems

The process of induction involves the following steps.

Step 1 :

Verify that the statement is true for n = 1, that is, verify that P(1) is true. This is a kind to climbing the first step of the staircase and is referred to as the initial step.

Step 2 :

Verify that the statement is true for n = k + 1 whenever it is true for n = k, where k is a positive integer. This means that we need to prove that P(k + 1) is true whenever P(k) is true. This is referred to as the inductive step.

Step 3 :

If steps 1 and 2 have been established then the statement P(n) is true for all positive integers n.

Mathematical Induction Problems With Solutions

Question 1 :

By the principle of mathematical induction, prove that, for n ≥ 1

13 + 23 + 33 + · · · + n3 = [n(n + 1)/2]2

Solution :

Let p(n) =  13 + 23 + 33 + · · · + n3 = [n(n + 1)/2]2

Step 1 :

put n = 1

p(1)  = 13 + 23 + 33 + · · · + 13 = [1(1 + 1)/2]2

  1  =  1

Hence p(1) is true.

Step 2 :

Let us assume that the statement is true for n = k

p(k)  =  13 + 23 + 33 + · · · + k3 = [k(k + 1)/2] -----(1)

We need to show that P(k + 1) is true. Consider,

Step 3 :

Let us assume that the statement is true for n = k + 1

p(k + 1)  =  13 + 23 + 33 + · · · + (k + 1)3 = [(k + 1)(k + 2)/2]2

 13 + 23 + 33 + · · ·k3 + (k + 1)3 = [(k + 1)(k + 2)/2]2

By applying (1) in this step, we get

(k + 1)2 (k2 + 4k + 4)/4  =  [(k + 1)(k + 2)/2]2

(k + 1)2 (k + 2)2 /4  =  [(k + 1)(k + 2)/2]2

By taking square for the entire terms, we get

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

Hence, by the principle of mathematical induction, for n≥1 

13 + 23 + 33 + · · · + n3 = [n(n + 1)/2]2

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