Use mathematical induction to prove the statement is true for all positive integers n.

10 + 30 + 60 + ... + 10n = 5n (n + 1)

F (n) = 10 + 30 + 60 + ... + 10n = 5n (n + 1)

1. Let n = 1.

LHS = 10n = 10 * 1 = 10 [ LHS - Left hand side ]

RHS = 5n (n + 1) = 5*1*2 = 10 [ RHS - Right hand side ]

LHS = RHS

Hence, f (n) is valid for n = 1.

2. Supposing that f (n) is valid for m.

So,

10 + 30 + 60 + ... + 10m = 5m (m + 1)

3. Let n = m + 1

LHS = 10 + 30 + 60 + ... + 10 (m + 1) = 10 + 30 + 60 + ... + 10m + 10

RHS = 5 (m + 1) (m + 1 + 1) = 5 (m + 1) (m + 2) = 5m (m + 1) + 10

We know from step 2, that 10 + 30 + 60 + ... + 10m = 5m (m + 1).

Hence,

10 + 30 + 60 + ... + 10m + 10 = 5m (m + 1) + 10

LHS = RHS

Hence, f (n) is valid for n = m + 1

Thus,

f (n) is valid for all positive integers