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The sum of n terms of a geometric sequence is given as a*(r^n - 1)/(r - 1), where r is the common ratio and a is the first term
4^n - 1
=> (2^(2n) - 1) / (2 - 1)
Here a = 1 and r = 2. But 2n can only take on even values.
The expression 4^n - 1 can be the sum of terms of a geometric sequence but it holds only when the number of terms is even.
We'll establish a numerical expression for the sum of n terms:
S, = (4^n)-1
We'll determine the general term bn, and then, we'll utter any other term of the progression.
bn=(4^n) - 1 - 4^(n-1) + 1
We'll eliminate like terms and we'll factorize by 4^n:
Since we know the general term bn, we'll compute the first 3 consecutive terms, b1,b2,b3.
Following the rule of a geometric sequence, we'll verify if
Sn = 4^n - 1 is the sum of the n terms of a geometric sequence.
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