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.
Question 1 :
By the principle of mathematical induction, prove that, for n ≥ 1
1^{3} + 2^{3} + 3^{3} + · · · + n^{3} = [n(n + 1)/2]^{2}
Solution :
Let p(n) = 1^{3} + 2^{3} + 3^{3} + · · · + n^{3} = [n(n + 1)/2]^{2}
Step 1 :
put n = 1
p(1) = 1^{3} + 2^{3} + 3^{3} + · · · + 1^{3} = [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) = 1^{3} + 2^{3} + 3^{3} + · · · + k^{3} = [k(k + 1)/2]^{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) = 1^{3} + 2^{3} + 3^{3} + · · · + (k + 1)^{3} = [(k + 1)(k + 2)/2]^{2}
1^{3} + 2^{3} + 3^{3} + · · ·k^{3} + (k + 1)^{3} = [(k + 1)(k + 2)/2]^{2}
By applying (1) in this step, we get
(k + 1)^{2} (k^{2} + 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
1^{3} + 2^{3} + 3^{3} + · · · + n^{3} = [n(n + 1)/2]^{2}
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