Question

rewrite in expanded form

Answer 1

\[\sum_{m=1}^{5} (2m^2+3)\]

Answer 2

thats the answer though

Answer 3

you need to sum each term for m=1, then m=2, then m=3, ... up to m=5

Answer 4

so, for example, the first term would be for m=1, giving:\[2(1)^2+3=2+3=5\]

Answer 5

samm2121: do you understand?

Answer 6

when i write my answer i write it like this (2m+3)+(2m^+3)+(2m^3+3)+(2m^4+3)+(2m^5+3)

Answer 7

each term of the sum is of the form \((2m^2+3)\)

Answer 8

and m goes from 1 to 5

Answer 9

asnaseer would you mind giving me the answer please

Answer 10

follow asnaseer's solution... I just realised my mistake

Answer 11

I can give you a hint instead. the first two terms would be:\[(2(1)^2+3) + (2(2)^2+3)\]

Answer 12

repeat this pattern for all 5 terms

Answer 13

okay would this be the answer 7+16+21+35+53 ?

Answer 14

how do you get 7 for the first term?
if you look further up I worked out the first term for you as follows:\[2(1)^2+3=2+3=5\]

Answer 15

sorry i dont get it

Answer 16

what part don't you get?

Answer 17

sorry i just dont get how its not 7 i added and did as the formula follows

Answer 18

ok, lets take a look at the first term in detail:\[2(1)^2+3=2+3=5\]do you agree that \((1)^2=1\)?

Answer 19

\[2\times 1^2 + 3 = 5\]

Answer 20

yes

Answer 21

so then we have:\[2(1)^2+3=2*1+3=2+3=5\]

Answer 22

does that make sense now?

Answer 23

remember that:\[a(b)=a\times b\]

Answer 24

ok thanks

Answer 25

yw