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

1 Answer
Mar 27, 2018

You can use the following general formula:

sum_(k=1)^n(ak+b)=n/2[a(n+1)+2b]nk=1(ak+b)=n2[a(n+1)+2b]

where the series has nn terms aa and bb 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)nk=1(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)^nbnk=1(ak+b)=nk=1ak+nk=1b

We can factor the aa.

sum_(k=1)^n(ak+b)=asum_(k=1)^nk+sum_(k=1)^nbnk=1(ak+b)=ank=1k+nk=1b

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

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

and

sum_(k=1)^nb=nbnk=1b=nb

So

sum_(k=1)^n(ak+b)=(an(n+1))/2+nb=n/2[a(n+1)+2b]nk=1(ak+b)=an(n+1)2+nb=n2[a(n+1)+2b]

EXAMPLE

Suppose we wanted the following sum:

7+10+13+16+19+22+25+28+317+10+13+16+19+22+25+28+31

There are 9 terms in this series so n=9n=9. Consecutive terms differ by 3 so a=3a=3. The first term, x_1=7x1=7 and so b=x_1-a=7-3=4b=x1a=73=4.

We want to find

sum_(k=1)^9(3k+4)=9/2[3(9+1)+2*4]=1719k=1(3k+4)=92[3(9+1)+24]=171

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