Question

@amistre64

Answer 1

How do I evaluate the series

\[\sum_{n=3}^{8}5n\]

Answer 2

@hartnn

Answer 3

\(\sum_3^8 = \sum_1^8-\sum_1^2\)

Answer 4

5 is constant, it can be thrown out of summation

Answer 5

ok what does that mean exactly.

Answer 6

sum of numbers from 3 to 8 = sum of numbers from 1 to 8 - sum of numbers from 1 to 2

Answer 7

ok what would we do next

Answer 8

find those 2 summations using the formula,
\(\sum _1^nn = \dfrac{n(n+1)}{2}\)

for 1st summation, n = 8
for 2nd summation, n = 2
then subtract :)

Answer 9

ok uhm how would i do that I dont know where to put the numbers

Answer 10

\(\sum_3^8 n= \sum_1^8n-\sum_1^2n = \dfrac{8*(8+1)}{2}- \dfrac{2*(2+1)}{2} =...\)

Answer 11

ok well that gives us

(28/2)+(2/2) right?

Answer 12

i corrected myself...please try again

Answer 13

I switched the - to a + was there something else I needed to change?

Answer 14

8*9/2 - 2*3/2 = 72/2 -6/2

Answer 15

oh ok sorry so that would give us

36-3

Answer 16

ys.
and don't forget the 5 we thrown out,

5(36-3) =...

Answer 17

ok so that gives us 5*33

Answer 18

yesh, any doubts anywhere ?

Answer 19

no i dont think so

Answer 20

ccol, 165 is your final answer :)

Answer 21

thx so much now can you help with a few more I am having trouble with it