Question 105132
{{{S =1*2+2*2^2+3.2^2}}}{{{....}}}{{{100*2^100}}}

{{{2S = 1*2^2+2*2^3+......}}}{{{+99*2^100 + 100*2^ 101}}}

Subtracting these two,  we get:

{{{S}}} - {{{2S}}} = {{{-S}}}

{{{-S = 2+2^2+2^3}}}{{{...........}}}{{{100*2^101}}}

{{{S=100*2^101-2(2^100-1)/(2-1)}}}

= {{{100*2^101-2^101+2}}}

= {{{99*2^101+2}}}