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

The answer is 1 because you have to flip the sign when dividing by negative number and 3