Question

Solve :(

Answer 1

\[\sum_{n-1}^{5}(2-3n)\]

Answer 2

no wonder you have a sad face at the end of that question o.O

Answer 3

the big intimidating sigma is just a fancy plus sign
replace \(n\) by 1, put a plus sign, replace \(n\) by 2, put a plus sign etc all the way up to 5
then add

Answer 4

\[\sum_{n=1}^{5}(2-3n)\]
\[=2-3\times 1+2-3\times 2+2-3\times 3+2-3\times 4+2-3\times 5\] is what you need to compute

Answer 5

you can rearrange as \(5\times 2-3\times (1+2+3+4+5)\) to make the calculation easier

Answer 6

you ok from there?

Answer 7

I think so, the final is 105 right

Answer 8

oh no it will be negative for sure

Answer 9

oh. dang

Answer 10

\[5\times 2-3\times (1+2+3+4+5)\]
\[=10-3\times 15=10-45=-35\] i think is correct

Answer 11

oooh okay. thanks that's an answer for me lol :)

Answer 12

yw, hope steps are clear