How many terms of the series 3+6+12+24+...add to 765?

looks like a geometric series with


a = 3
r = 2

sum(n) = a(r^n -1)/(r-1)
765 = 3(2^n - 1)/1
255 = 2^n -1
256 = 2^n
2^8 = 2^n

n = 8

there are 8 terms

why didn't you use this equation: 765=3(2)^n-1

wait I understand thanks for your help :D

To find out how many terms of the series 3+6+12+24+... add to 765, we need to identify the pattern of the series and determine an equation to solve the problem.

In this series, each term is obtained by multiplying the previous term by 2. Starting with the first term, we have:
3, 6, 12, 24, ...

To find the general term of the series, we can use the formula for the nth term of a geometric progression:

aₙ = a₁ * r^(n-1),

where:
aₙ is the nth term of the series,
a₁ is the first term of the series,
r is the common ratio between the terms, and
n is the number of terms.

In this case:
a₁ = 3 (the first term),
r = 2 (common ratio).

We need to find the value of n for which the sum of the terms is equal to 765. To find this value, we can use the formula for the sum of the first n terms of a geometric progression:

Sₙ = a₁ * (1 - r^n) / (1 - r),

where:
Sₙ is the sum of the first n terms.

We can rearrange this formula to solve for n:

n = log(Sₙ * (1-r) / a₁ + 1) / log(r) + 1.

Let's substitute the known values into this equation:

n = log(765 * (1-2) / 3 + 1) / log(2) + 1.

Using a calculator, we can evaluate this expression to find the value of n, which will tell us how many terms of the series add up to 765.