# A geometric series has third term `36` and sixth term `972.` a) find the first and the common ratio of the series.I am able to solve this one which I got `a = 4` and `r = 3.` My problem is in...

A geometric series has third term `36` and sixth term `972.`

a) find the first and the common ratio of the series.

I am able to solve this one which I got `a = 4` and `r = 3.`

My problem is in b): the n-th term of the series is `U_n,`

(i) show that `sum_(n = 1)^20 U_n = K(3^(20) - 1)` where `K` is an integer to be found.

How do I show this?

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Hello!

I agree with your answer to the part **a**), the only possible series is `U_n = 4*3^(n-1).`

The question **b**) becomes simple if we recall the formula of the sum of `N` terms of a geometric progression `U_n` with the common ratio `r:`

`sum_(n = m)^(m+N) U_n = ((U_((m+N))) - (U_m))/(r-1).`

I give the more general form of the common formula because sometimes there is a confusion related with the starting index of the sum (0 or 1). In this from, the sum is

((the last summed up term of the series) - (the first summed up term of the series)) above (the common ratio - 1).

We know already that `U_n = 4*3^(n-1),` `m = 1` and `N = 20.`

Therefore `U_(m+N) = U_21 = 4*3^(21 - 1) = 4*3^20` and the sum is equal to

`(4*3^20 - 4*3^0)/(3 - 1) = 4/2 * (3^20 - 1) = 2*(3^20 - 1).`

Hence the statement we need to prove is true and **K = 2**. This is the answer.

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