find 18 "sigma notation" i=7 11i+3

Question

anonymous · Answer

18 is the upper limit
i=7 is the lover limit
and 11i+3 is next to the sigma notation

amistre64 · Answer

i changes from 7 to 18

  11 (7) + 3
+11 (8) + 3
+11(9) + 3
+11(10) + 3
+11(11) + 3
....

amistre64 · Answer

sum (from i=7 to 18) of f(i)  is a good way to describe it

anonymous · Answer

so I do that till i get to 18?

anonymous · Answer

what do I do after that?

amistre64 · Answer

that is the brute math way; yes

amistre64 · Answer

there are more elegant methods im sure

amistre64 · Answer

$$\sum_{i=7}^{18}11i+\sum_{i=7-6}^{18-6}3$$

anonymous · Answer

so wait Im trying to find the sum of all those numbers? 
So do I add them?

amistre64 · Answer

yes, sum means to add up all those values

anonymous · Answer

$$n/2(a+l)$$

amistre64 · Answer

$$11\sum_{i=7}^{18}i+3\sum_{i=1}^{12}$$
$$11\frac{12(7+18)}{2}+3(12)$$$$66(25)+36$$
$$1650+36=1686$$
if i did it right :)

anonymous · Answer

oh my gosh thankyou soo much! Thats an answer choice, and I think i get it a little better! You're the best!

amistre64 · Answer

just because its an answer choice, doesnt mean its right .... but yes, its right lol

anonymous · Answer

interesting...

anonymous · Answer

@amistre, now you have me thinking about how to prove:$$a_{1}+a_{2}...a_{n}=n \frac{(a_{n}+a_{1})}{2}$$in general.

anonymous · Answer

@amistre64

anonymous · Answer

Ok, after pondering this one, it isn't true in general.  It is true when the terms of the series are related linearly.

amistre64 · Answer

:)  its true of an arithmetic progression ... if im remembering correctly

amistre64 · Answer

or as you put it, a linear relationship ... yes

anonymous · Answer

Yeah, i'm very rusty with sequences and series...still plugging away at this idea :)

anonymous · Answer

what is a linear relationship?

anonymous · Answer

A linear relationship has the form:$$y=mx+b$$ where m and b are constants.