If `t_(1)=1,t_(r )-t_( r-1)=2^(r-1),r ge 2`, find `sum_(r=1)^(n)t_(r )`.

Explanation

`t_(1)=1 " and "t_(r )-t_( r-1)=2^(r-1),r ge 2`
`t_(2)-t_(1)=2`
`t_(3)-t_(2)=2^(2)`
`t_(4)-t_(3)=2^(3)`
` vdots " " vdots " " vdots `
`t_(n)-t_(n-1)=2^(n-1)`
Adding conlumnwise, we get
`t_(n)-t_(1)=2+2^(2)+"........."+2^(n-1)`
`t_(n)=1+2+2^(2)+"......"+2^(n-1)`
`t_(n)=(1*(2^(n-1)))/(2-1) implies t_(n)=2^(n)-1`
So,`sum_(r=1)^(n)t_(r )=t_(1)+t_(2)+"........"+t_(n)=(2-1)+(2^(2)-1)+"......"+(2^(n)-1)`
`=(2+2^(2)+"......"+2^(n))-n=(2*(2^(n)-1))/((2-1))-n=2^(n+1)-2-n`
`=2^(n+1)-n-2`.