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.
- print Print
- list Cite
Expert Answers
calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
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
=>...
(The entire section contains 85 words.)
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Related Questions
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!
Student Answers