Question

Using mathematical induction for integers n `>=` 1, prove a + (a+d) +...+ (a+nd) = 1/2 (n+1)(2a+nd)?

Answer 1

Claim: `sum_(i=0)^n(a+id)=((n+1)(2a+nd))/2` for all integers `n>=1`

Proof (by induction):

Base case:
If `n=1`, then

`sum_(i=0)^n(a+id) = 2a+d = (2(2a+d))/2 = ((n+1)(2a+nd))/2`

Inductive hypothesis:
Suppose that the claim holds true for some integer `k>=1`.

Induction step:
We wish to show that the claim holds for `k+1`. Indeed,

`sum_(i=0)^(k+1)(a+id) = a+(k+1)d + sum_(i=0)^k(a+id)`

`=a+(k+1)d + ((k+1)(2a+kd))/2`
(by the inductive hypothesis)

`=(2a+2(k+1)d)/2+((k+1)(2a+kd))/2`

`=(2a+2(k+1)d+(k+1)(2a+kd))/2`

`=(2a+2(k+1)d+2(k+1)a+k(k+1)d)/2`

`=(2(k+2)a+(k+1)(k+2)d)/2`

`=((k+2)(2a+(k+1)d))/2`

`=(((k+1)+1)(2a+(k+1)d))/2`

as desired.

We have supposed the claim is true for `k` and shown it true for `k+1`, thus, by induction, it is true for all integers `n>=1`.
∎