Question

Find the sum of this series: 3^0+3^1+3^2+3^3...+3^100

Answer 1

This is a geometric progression.

Answer 2

sum = 3(1-3^100)/(1-3) +1 = 7.73x10^47

Answer 3

Plz explain

Answer 4

Hmmm.....that formula can't be true for "r>1".

Answer 5

Technically, \( 1.(3^{101} -1)/2\)

Answer 6

How?

Answer 7

i think the formula works..

Answer 8

Yeah its one of my answer choices

Answer 9

But explain how you got it plz

Answer 10

@Callisto: It works here but not in general. Try taking limit where \( n\to \infty \)

Answer 11

Isnt limit 100?

Answer 12

s = 1 + x + x^2 + ... + x^n
xsn = x + x^2 + x^3 + ... + x^(n+1)

xs - s = x^(n+1) - 1

s = (x^(n+1) -1)/(x-1)

Answer 13

then that would be GS sum to infinity that would be sum = a(1-r)/r

Answer 14

Thanks

Answer 15

@Callisto - you keep using the formula that works for r<1.

Answer 16

slaaibak has it right.

Answer 17

Thanks everybody medals for everyone :D

Answer 18

sorry that should be a/(1-r)

Answer 19

for me r<1 or r>1 is the same, just changing 1-r^n to r^n-1 and change the base also.. that's means taking out -1 as the common factor lol

Answer 20

@Callisto: YOu will see the diff when you take the limit

Answer 21

\[\lim_{n\to inf}\frac{1-r^n}{1-r}=\frac{-inf}{1-r}=-\frac{inf}{1-r}\]

\[\lim_{n\to inf}\frac{r^n-1}{r-1}=\frac{inf}{r-1}=\frac{inf}{-(1-r)}=-\frac{inf}{1-r}\]

so were am I missing this?

Answer 22

read the second post also

Answer 23

http://www.wolframalpha.com/input/?i=limit+to+inf+%281-4%5En%29%2F%281-4%29

http://www.wolframalpha.com/input/?i=limit+%284%5En-1%29%2F%284-1%29+to+inf

Answer 24

sum = 3(1-3^100)/(1-3) +1 = 7.73x10^47

this one?

Answer 25

A = 3^0+3^1+3^2+3^3+...+3^100
-3A = -3^1-3^2-3^3- ...-3^100-3^101
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(1-3)A = 3^0 -3^101

A = 3^0-3^101
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