Question

CAN SOMEONE PLEASE HELP i only need help with the second one in the picture.

Answer 1

didnt you have tons of replies for this one?

Answer 2

nope.

Answer 3

u mean B. right?

Answer 4

yeah i did the first one already

Answer 5

there r 8 terms.
the first term is for n=1 so it is
\[ \frac{2\cdot1}{3}=\frac{2}{3}. \]
the last term is for n=8 so it is
\[ \frac{2\cdot8}{3}=\frac{16}{3} \]

Answer 6

so the amount of terms is always just the number on top?

Answer 7

for the last part
\[\large \sum_{n=1}^8\frac{2n}{3}=\frac{2}{3}\sum_{n=1}^8n=
\frac{2}{3}\cdot\frac{8(8+1)}{2}=\frac{8\cdot9}{3}=24 \]

Answer 8

NO. if u have
\[\large \sum_{n=a}^A \]
then the number of terms is A-a+1

Answer 9

thank you so much

Answer 10

u r welcome