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

Dec 11, 2016

Claim: ${\sum}_{i = 0}^{n} \left(a + i d\right) = \frac{\left(n + 1\right) \left(2 a + n d\right)}{2}$ for all integers $n \ge 1$

Proof (by induction):

Base case:
If $n = 1$, then

${\sum}_{i = 0}^{n} \left(a + i d\right) = 2 a + d = \frac{2 \left(2 a + d\right)}{2} = \frac{\left(n + 1\right) \left(2 a + n d\right)}{2}$

Inductive hypothesis:
Suppose that the claim holds true for some integer $k \ge 1$.

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

${\sum}_{i = 0}^{k + 1} \left(a + i d\right) = a + \left(k + 1\right) d + {\sum}_{i = 0}^{k} \left(a + i d\right)$

$= a + \left(k + 1\right) d + \frac{\left(k + 1\right) \left(2 a + k d\right)}{2}$
(by the inductive hypothesis)

$= \frac{2 a + 2 \left(k + 1\right) d}{2} + \frac{\left(k + 1\right) \left(2 a + k d\right)}{2}$

$= \frac{2 a + 2 \left(k + 1\right) d + \left(k + 1\right) \left(2 a + k d\right)}{2}$

$= \frac{2 a + 2 \left(k + 1\right) d + 2 \left(k + 1\right) a + k \left(k + 1\right) d}{2}$

$= \frac{2 \left(k + 2\right) a + \left(k + 1\right) \left(k + 2\right) d}{2}$

$= \frac{\left(k + 2\right) \left(2 a + \left(k + 1\right) d\right)}{2}$

$= \frac{\left(\left(k + 1\right) + 1\right) \left(2 a + \left(k + 1\right) d\right)}{2}$

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 $n \ge 1$.