令S=1+2·3+3·(3^2)+4·(3^3)+……+100·(3^99)
则3S= 3+2·(3^2)+3·(3^3)+……+99·(3^99)+100·(3^100)
上面两式相减得：
S-3S=1+3+3^2+3^3+……+3^99-100·(3^100)
-2S=(1-3^100)/(1-3)-100·(3^100)
=(3^100-1)/2-100·(3^100)
=-199/2·(3^100)-1/2
所以S=199/4·(3^100)+1/4
即1+2·3+3·(3^2)+4·(3^3)+……+100·(3^99)=199/4·(3^100)+1/4