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

##### 1 Answer

**Claim:**

**Proof (by induction):**

**Base case:**

If

**Inductive hypothesis:**

Suppose that the claim holds true for some integer

**Induction step:**

We wish to show that the claim holds for

#=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

∎