Question

If $t_1=1,t_r-t_{r-1}=2^{r-1},r\ge 2$. Then find $\sum _{r=1}^n{t}_r$ for $n=3$?

Answer 1

$t_1=1$

$t_r-t_{r-1}=2^{r-1}(r\ge 2)$

At $r=2,t_2-t_{2-1}=2^{2-1}$

$\Rightarrow t_2-t_1=2$

$\Rightarrow t_2-1=2$

$\Rightarrow t_2=3$

At $r=3,t_3-t_{3-1}=2^{3-1}$

$\Rightarrow t_3-t_2=4$

$\Rightarrow t_3-3=4$

$\Rightarrow t_3=7$

$S=\sum _{r=1}^n{t}_r$ for $n=3$

$\Rightarrow S=t_1+t_2+t_3=1+3+7=11$