Question 1036497
The alpha sequence is an arithmetic sequence with first term 10 and common difference -2, and is defined by the formula {{{a[n] = 12-2n = 2(6-n)}}} for n = 1,2,3,... .

The beta sequence is a geometric sequence with first term 9 and common ratio 2/3, and is defined by the formula {{{b[n] = 9*(2/3)^(n-1)}}}.

The gamma sequence is thus defined by the formula {{{g[n] = 2(6-n)* 9*(2/3)^(n-1) = 18(6-n)*(2/3)^(n-1) }}}.

We have to find {{{sum(18(6-n)*(2/3)^(n-1),n = 1, infinity)}}}

= {{{18sum((6*(2/3)^(n-1)-n*(2/3)^(n-1)),n = 1, infinity)}}}

= {{{18sum(6*(2/3)^(n-1),n = 1, infinity)-18sum(n*(2/3)^(n-1), n=1, infinity)}}}

= {{{108sum((2/3)^(n-1),n = 1, infinity)-18sum(n*(2/3)^(n-1), n=1, infinity)}}}

The infinite geometric series {{{1+x+x^2+x^3}}}+... = {{{1/(1-x)}}} as long as -1 < x < 1.
Staying within the domain of convergence and differentiating term-by-term, we get
{{{1+2x+3x^2+4x^3}}}+... = {{{1/(1-x)^2}}}


Now the first infinite sum is an infinite geometric series with sum {{{1/(1-2/3) = 1/(1/3) = 3}}}.
The second infinite sum is derivative of the infinite geometric series and has sum {{{1/(1-2/3)^2 = 1/(1/3)^2 = 1/(1/9) = 9}}}.

Therefore...

{{{108sum((2/3)^(n-1),n = 1, infinity)-18sum(n*(2/3)^(n-1), n=1, infinity)}}}

= 108*3 - 18*9 = {{{highlight(162)}}}.