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 #
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check:
Tony B