The sum of the series 1+2⋅2+3⋅22+4⋅23+5⋅24+…+100⋅299 is

Let S=1+2⋅2+3⋅22+4⋅23+5⋅24 +…+100⋅299∴ 2S=1⋅2+2⋅22+3⋅23+…+99⋅299  +100⋅2100 Substracting, we get −S=1+1⋅2+1⋅22+…+1⋅299−100⋅2100=1+2+22+…+299−100⋅2100=12100−12−1−100⋅2100=2100−1−100⋅2100∴S=100⋅2100−2100+1=99⋅2100+1.