Question 401421
I solved a problem very similar to this one. The infinite series can be expressed as (1+p+p^2+...) + (p+p^2+p^3+...) + (p^2+p^3+p^4+...) = (1+p+p^2+p^3+...)(1+p+p^2+p^3+...) = {{{(sum(p^i, i = 0, infinity))^2}}}. The sum of a geometric series {{{sum(p^i, i = 0, infinity)}}} is given by {{{1/(1-p) = (1-p)^(-1)}}}, so squaring it, we obtain {{{(1-p)^-2}}}. Note that {{{-1 < p < 1}}} for this to occur, as the series diverges otherwise.