Induction: when $a_k=a_0+(k-1)d$ for any non-negative integer $k$, prove $\sum_{i=1}^n a_i=\frac{n(a_1+a_n)}{2}$

Question

Let $a_k = a_0 + (k-1)d$ for any non-negative integer $k$, prove by mathematical induction that

$$\sum_{i=1}^n a_i = \frac{n(a_1+a_n)}{2}$$

My plan so far is to

1. Produce the iteration

$$a_1 = a_0 +(1-1)d,\qquad a_2 = a_0 +(2-1)d, \qquad \cdots, \qquad a_n = a_0 +(n-1)d,$$

Problem: can't find $a_0$.

2. Sub lowest base value 1

$$\frac{n\big( (a_0 +(1-1)d) + (a_0 +(n-1)d) \big)}{2} = \frac{n(a_0 d + a_0 + nd -d)}{2}$$

Not sure what I can do with this value here and am stuck.

Please, if possible, leave the answers in the simpliest form so I can understand.

If you initial term is $a_1$, you should have $a_k = a_1 + (k - 1)d$. Then $a_1 = a_1 + (1 - 1)d = a_1 + 0d = a_1$, as expected. — N. F. Taussig, Sep 14 '19 at 13:51

score 1 · Answer 1 · answered Sep 14 '19 at 13:32

Gauss was famous for solving this as a child. Write the sum twice, with a reversed version under the original. You get $n$ columns each summing to $a_1+a_n$. (The inductive part is verifying that, as you move to the right, each column has this sum.)

Sangchul Lee · Accepted Answer · 2019-09-14T13:40:19.437

Here is a famous trick allegedly attributed to Gauss: Let $i, j$ satisfy $i+j=n+1$. Then

\begin{align*} a_i + a_j &= (a_0 + (i-1)d) + (a_0 + (j-1)d) \\ &= 2a_0 + (i+j-2)d \\ &= 2a_0 + (n-1)d \\ &= a_1 + a_n. \end{align*}

Now, by reversing the order of summation, we have $\sum_{i=1}^{n} a_i = \sum_{i=1}^{n} a_{n+1-i}$, and so,

\begin{align*} 2 \sum_{i=1}^{n} a_i &= \left( \sum_{i=1}^{n} a_i \right) + \left( \sum_{i=1}^{n} a_{n+1-i} \right) \\ &= \sum_{i=1}^{n} (a_i + a_{n+1-i}) \\ &= \sum_{i=1}^{n} (a_1 + a_n) \\ &= n(a_1 + a_n). \end{align*}

Dividing both sides by $2$ then proves the desired identity. Here is a visualization of this trick for $n = 10$:

Induction: when $a_k=a_0+(k-1)d$ for any non-negative integer $k$, prove $\sum_{i=1}^n a_i=\frac{n(a_1+a_n)}{2}$

2 Answers2