Question

Find the 6th partial sum of
a 78,432
b 2,387
c 9,555
d 11,204

Answer 1

\[\sum_{i=1}^{\infty} 7(4)^{i-1}\]

Answer 2

factor out the constants

Answer 3

7/4 (4^1+4^2+...) which is a geometric sum if anything right?

Answer 4

An \(n\)th partial sum is the evaluation of the sum to \(n\) terms.

So, evaluate
\[\sum_{1 \leq i \leq 6}7(4)^{i-1}\]

Answer 5

if we go ahead and factor out another 4 to make the series start at 4^0 we get:

7(4^0 + ... 4^5) i believe

Answer 6

I have no idea what you are saying

Answer 7

\[7(4^0)+7(4^1)+7(4^2)+7(4^3)+7(4^4)+7(4^5)\]
\[7(1+4+16+...+4^5)\]

Answer 8

Okay amistre64 why am I doing this? I know what you are doing, just not why

Answer 9

im trying to simplify it out to get to the geometric part that i can use the formula on

Answer 10

Okay

Answer 11

hence:\[7*\frac{1-4^5}{1-4}\]

if i recall it right

Answer 12

Okay, then what?

Answer 13

or is it 4^6 up there? hmm

Answer 14

It should be 4^5 because i am looking for the sixth. right

Answer 15

1 4 4^2 ... 4^5
-(4 4^2 ... 4^5 4^6)

im going with 4^6

Answer 16

16*16*16

need a calculator lol

Answer 17

That is 4096

Answer 18

\[
\sum_{1 \leq k \leq n}ar^k=a\frac{1-r^{n}}{1-r}
\]

\[
\sum_{1 \leq i \leq 6}7(4)^{i-1}=\sum_{0 \leq i \leq 5}7(4)^i=7\frac{1-4^5}{1-4}
\]

Answer 19

AHHHH! I am so confused!!

Answer 20

-1 and divide by 3 then

Answer 21

i get 9555 over all

Answer 22

Limitless, what did you get?

Answer 23

http://www.wolframalpha.com/input/?i=sum+7*%284%5En%29%2C+n%3D0+to+5

yay!!

Answer 24

Okay, because Limitless is gone now, I guess I am going with 9555

Answer 25

Apparently the geometric sum formula is wrong. It should be
\[
\sum_{0 \leq i \leq n}ar^i=\frac{a(k^{n+1}-1)}{k-1}
\]
So:
\[
\sum_{0 \leq i \leq 5}7(4)^i=\frac{7(4^{5+1}-1)}{4-1}=\frac{7(4^{6}-1)}{3}=9,555\]

Answer 26

Okay, thanks guys!