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?

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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