Question

I really need help

Answer 1

on?

Answer 2

\[\sum_{k}^{12}=1^{(-2+6k)}\]

Answer 3

can you find the indicated sum?

Answer 4

=12

Answer 5

how did you get that?

Answer 6

I assume that k goes from 1 to 12

Answer 7

Find the indicated sum for the arithmetic series.

Answer 8

if k =1, then the exponent is -2+6 =4 , so that 1^4=1
if k =2 , then the exponent is -2+ 12 = 10 , so that 1^10 =1
until k =12, you have 12 times 1 = 12

Answer 9

Answer choices are
A:888
B:368
C:565
D:444

Answer 10

if they are options, then recheck your expression, 6k can't be in the exponent

Answer 11

\[\sum_{k=1}^{12}1^{(-2+6k)}= 1^{(-2+6*1)}+1^{(-2+6*2)}+........+1^{(-2+6*12)}=12\]

Answer 12

hold on let me draw it