Asked by ABBY
How do prove without using a formula , that the sum of the series 3²+3³+3to the power of 6+...to 20 terms is given by S20 =9/8(9to the power of 20 -1)
Answers
Answered by
Reiny
You are basically going through the development of the formula, using a specific example
You seem to have a typo
your: 3²+3³+3to the power of 6+...to 20 terms
makes little sense
I will assume we have
S(20) = 3^2 + 3^3 + ... + 3^21
multiply both sides by 3
3S(20) = 3^3 + 3^4+ ... + 3^22
subtract them
S(20) - 3(s(20) = 3^2 - 3^22 , all terms in between dropped out
S(20)(1-3) = 3^2( 1 - 3^20)
S(20) = 3^2(3^20 -1)/2
= (9/2)(3^20 - 1)
check: for my answer using the formula
a = 3^2 or 9
r = 3
Sum(20) = 9(3^20 - 1)/(3-1)
= (9/2)(3^20 - 1) , which is what I have
which is different from the answer you have by a huge factor
Check your typing.
You seem to have a typo
your: 3²+3³+3to the power of 6+...to 20 terms
makes little sense
I will assume we have
S(20) = 3^2 + 3^3 + ... + 3^21
multiply both sides by 3
3S(20) = 3^3 + 3^4+ ... + 3^22
subtract them
S(20) - 3(s(20) = 3^2 - 3^22 , all terms in between dropped out
S(20)(1-3) = 3^2( 1 - 3^20)
S(20) = 3^2(3^20 -1)/2
= (9/2)(3^20 - 1)
check: for my answer using the formula
a = 3^2 or 9
r = 3
Sum(20) = 9(3^20 - 1)/(3-1)
= (9/2)(3^20 - 1) , which is what I have
which is different from the answer you have by a huge factor
Check your typing.
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