Question

Sum of the terms of a geometric sequenceVerify if the sum 4^n - 1 represents the sum of the terms of a geometric sequence.

Accepted Answer

Let 4^n - 1 represent the sum of m terms of a GP. 4^(n - 1) - 1 is the sum of m - 1 terms of the GP.
Tn = 4^n - 1 - 4^(n - 1) + 1
=> 4^n - 4^n/4
For two consecutive terms Tn and Tn+1
Tn+1 / Tn = [4^(n + 1) - 4^(n + 1)/4]/(4^n - 4^n/4)
=> [4*4^n - 4*4^n/4]/(4^n - 4^n/4)
=> 4*(4^n - 4^n/4)/(4^n - 4^n/4)
=> 4
As we arrive at a common ratio between consecutive terms, 4^n - 1 represents the sum of terms of a GP.

Answer

Finding the general term bn of the sequence, allows us to determine any term of the progression.
From enunciation:
Sn=b1+b2+b3+...+bn
(4^n)-1=b1+b2+b3+...+bn
bn=(4^n)-1-[b1+b2+b3+...+b(n-1)]
But [b1+b2+b3+...+b(n-1)]=S(n-1)=[4^(n-1)]-1
bn=(4^n) - 1 - 4^(n-1) + 1
We'll eliminate like terms and we'll factorize by 4^n:
bn=4^n(1-1/4)=4^n*3/4=3*4^(n-1)
Now, we'll calculate the first 3 consecutive terms, b1,b2,b3.
b1=3*4^0
b2=3*4^(2-1)=3*4
b3=3*4^(3-1)=3*4^2
According to  the rule of a geometric sequence, we'll verify if
b2=sqrt (b1*b3)
3*4= sqrt(3*1*3*16)
3*4=3*4
Sn = 4^n - 1 represents the sum of the terms of a geometrical progression!

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Sum of the terms of a geometric sequence Verify if the sum 4^n - 1 represents the sum of the terms of a geometric sequence.

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