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Question
Prove by method of induction, for all n ∈ N:
2 + 4 + 6 + ..... + 2n = n (n+1)
Solution
Let P(n) ≡ 2 + 4 + 6 + …… + 2n = n(n + 1), for all n ∈ N
Step I:
Put n = 1
L.H.S. = 2
R.H.S. = 1(1 + 1) = 2 = L.H.S.
∴ P(n) is true for n = 1.
Step II:
Let us consider that P(n) is true for n = k
∴ 2 + 4 + 6 + ……. + 2k = k(k + 1) …(i)
Step III:
We have to prove that P(n) is true for n = k + 1 i.e., to prove that
2 + 4 + 6 + …. + 2(k + 1) = (k + 1) (k + 2)
L.H.S. = 2 + 4 + 6 + …… + 2 (k + 1)
= 2 + 4 + 6 + …… + 2k + 2(k + 1)
= k(k + 1) + 2(k + 1) …[From (i)]
= (k + 1).(k + 2)
= R.H.S.
∴ P(n) is true for n = k + 1
Step IV:
From all steps above by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 2 + 4 + 6 + …… + 2n = n (n + 1) for all n ∈ N
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