How do you find the sum of a finite geometric sequence from n = 1 to n = 8, using the expression $−2(3)^n − 1$ ?

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Answer 1 · 2016-12-19T20:37:19

I could just use the formula, plug in the values and get the answer. Not going to do that!

The sum is $" " - 19688$

So that we can link this to the generic formula bit let the geometric ratio of 3 be $r$

Then we have:
$sum_(i=1->n) [-2r^i-1]$

$sum_(i=1->n)-2r^i-sum_(i=1->n)1$

but $sum_(i=1->n)1 = n$ giving

$sum_(i=1->n)[-2r^i] -n$

$-{ 2sum_(i=1->n)[r^i] +n }............ Expression 1$
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Set $" "t=sum_(i=1->n)[r^i]$

$" "t=r^1+r^2+r^3+...+r^n$

$" "rt=r^2+r^3+...+r^n+r^(n+1)$

$rt-t=r^(n+1)-r$

$t=(r(r^n-1))/(r-1)$
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Thus by substitution in $Expression(1)$ we have

$-{ 2sum_(i=1->n)[r^i] +n }" "-> -{color(white)(2/2) 2t +n }$

$(-1)xx{2xx(3(3^8-1))/(3-1)+8}$

$(-1)xx{cancel(2)^1xx(3(6560))/(cancel(2)^1)+8}$

$= - 19688$
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Check:
Tony B

How do you find the sum of a finite geometric sequence from n = 1 to n = 8, using the expression −2(3)^n − 1−2(3)^n − 1?