Question

Find the sum of the geometric sequence.
(3/2),(3/8),(3/32),(3/128),(3/512)
PLEASE HELP !
I GIVE MEDALS

Answer 1

Just add up the fractions... Or terms.

Answer 2

THats not the answer

Answer 3

@amoodarya and ideas ?

Answer 4

s=3/2+3/8+3/32+3/128 +3/512 +...
miltiply by 1/4 (r=1/4)
1/4s= 3/8+3/32+3/128 +3/512 +...

now s-1/4s=3/2
so 3/4s=3/2

Answer 5

I dont see that ..... these are the options @amoodarya

Answer 6

What did I say? Add up the fractions! That's the sum! the final result after adding is the sum! So the answer is C.

Answer 7

what about
Write the sum using summation notation, assuming the suggested pattern continues.

16 + 25 + 36 + 49 + ... + n2 + ...
@Zelda

Answer 8

1^2+2^2+3^2+.....n^2=n(n+1)(2n+1)/6
so 4^2+5^2+6^2+....n^2=1^2+2^2+3^2+.....n^2-(1+4+9)=n(n+1)(2n+1)/6-(1+4+9)

Answer 9

so which option ? @amoodarya

Answer 10

and for which question ?

Answer 11

s=3/2+3/8+3/32+3/128 +3/512
s=3/512(256+64+16+4+1)
s=1023/512

Answer 12

can you help me with that second one i just commented

Answer 13

@amoodarya

Answer 14

second question
did you mean 4^2+5^2+...n^2= ?
or 4^2+5^2+...n^2+....= ?

Answer 15

the second one @amoodarya

Answer 16

yeah the second question

Answer 17

4^2+5^2+...n^2+....= @amoodarya

Answer 18

for this question 16 + 25 + 36 + 49 + ... + n2 + ...

Answer 19

1^2+2^2+3^2+.....n^2+... =infinity
but
1^2+2^2+3^2+.....n^2=n(n+1)(2n+1)/6

so
4^2+5^2+6^2+....n^2=1^2+2^2+3^2+.....n^2- (1+4+9)=n(n+1)(2n+1)/6- (1+4+9)

Answer 20

The options are Im just having a hard time reading that

Answer 21

@amoodarya

Answer 22

D.