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. = R.H.S.
∴ P(n) is true for n = 1.
Step II: Let us assume 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 the 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.
