Question

How do I find the sum of an arithmetic sequence on a calculator?

Answer 1

You can use the following general formula:

`sum_(k=1)^n(ak+b)=n/2[a(n+1)+2b]`

where the series has `n` terms `a` and `b` are constants.

(Ask a general question, get a general answer.)

Explanation:

Seems that it would be handy to have a closed form expression for the following:

`sum_(k=1)^n(ak+b)`

where a and b are constants.

The commutative property of addition says

`sum_(k=1)^n(ak+b)=sum_(k=1)^nak+sum_(k=1)^nb`

We can factor the `a`.

`sum_(k=1)^n(ak+b)=asum_(k=1)^nk+sum_(k=1)^nb`

But we know closed form representations for both of these sums.

`sum_(k=1)^nk=(n(n+1))/2`

and

`sum_(k=1)^nb=nb`

So

`sum_(k=1)^n(ak+b)=(an(n+1))/2+nb=n/2[a(n+1)+2b]`

EXAMPLE

Suppose we wanted the following sum:

`7+10+13+16+19+22+25+28+31`

There are 9 terms in this series so `n=9`. Consecutive terms differ by 3 so `a=3`. The first term, `x_1=7` and so `b=x_1-a=7-3=4`.

We want to find

`sum_(k=1)^9(3k+4)=9/2[3(9+1)+2*4]=171`

which is, in fact, the sum of the numbers given.