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