Sum to n terms the series `1+3+7+15+31+...`

Explanation

The sequence of differences berween successive terms is `2,4,8,16,"…."`. Clearly, it is a GP with common ratio 2. So, let the nth term and sum of the series upto n terms of the series be `T_(n)` and `S_(n)`, respectively. Then,
`S_(n)=1+3+7+15+31+"..."T_(n-1)+T_(n) "....(i)"`
`S_(n)=1+3+7+15+"..."T_(n-1)+T_(n) "....(ii)"`
Subtracting Eq.(ii) from Eq.(i), we get
`0=1+2+4+8+16+"..."(T_(n)-T_(n-1))-T_(n)`
` implies T_(n)=1+2+4+8+16+"..." " upto n terms "`
` =(1*(2^(n)-1))/(2-1)`
Hence, `T_(n)=(2^(n)-1)`
`therefore` Sum of n terms `S_(n)=sumT_(n)=sum (2^(n)-1)=sum2^(n)-sum 1`
`=(2+2^(2)+2^(3)+"...."+2^(n))-n`
`=(2*(2-1))/(2-1)-n=2^(n+1)-2-n`