# how do you write the series 3/4-1/2-3-....-81 3/4 with sigma notation? :the general term is  (t(n)=3/4-5n(n-1)/8)

embizze | Certified Educator

We are given the general term for a series: `t(n)=3/4-(5n(n-1))/8` .

We are asked to write the series with sigma notation with this general term going from `3/4` to `-81 3/4` .

(1) The underlying sequence is `3/4,-1/2,-3,-6 3/4,-11 3/4,-18,-25 1/2,-34 1/4,-44 1/4,-55 1/2,-68,-81 3/4`

This sequence has 12 terms -- the series we are interested in also has 12 terms.

** Alternatively, and better if the series had been longer:

The last term is `-81 3/4` , so set `t(n)=-81 3/4` and solve for `n` . Here we get:

`-81 3/4=3/4 - (5n(n-1))/8`

`(5n(n-1))/8=82 1/2`

`5n(n-1)=660`

`5n^2-5n-660=0`

`5(n^2-n-132)=0`

`5(n-12)(n+11)=0`

Since `n>0` we have `n=12` ; thus there are 12 terms.

(2) The general form for sigma notation is `sum_(i=a)^b f(i)` , where a is the lower limit, b the upper limit, and f(i) the general term. We will start at 1, end at 12, and use the general form given.

(3) Putting it all together we get:

`sum_(i=1)^12 (3/4 - (5i(i-1))/8 )`