Could you please explain how come: sum_(i=1)^infty (0.5)^(i+1)(i+1)=2

Could you please explain how come:
$\sum _{i=1}^{\mathrm{\infty }}\left(0.5{\right)}^{i+1}\left(i+1\right)=2$
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Kennedy Evans
We know for $|r|<1$
$\sum _{1\le i<\mathrm{\infty }}{r}^{i+1}={r}^{2}\cdot \frac{1}{1-r}=\frac{{r}^{2}}{1-r}$

$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\sum _{1\le i<\mathrm{\infty }}\left(i+1\right){r}^{i}=\frac{2r-{r}^{2}}{\left(1-r{\right)}^{2}}$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\sum _{1\le i<\mathrm{\infty }}\left(i+1\right){r}^{i+1}=\frac{r\left(2r-{r}^{2}\right)}{\left(1-r{\right)}^{2}}$
Alternatively, using Arithmetico-geometric series, we can show,
$\sum _{1\le i<\mathrm{\infty }}\left(i+c\right){r}^{i+1}=\frac{\left(c+1\right){r}^{2}-c\cdot {r}^{3}}{\left(1-r{\right)}^{2}}$
where c is any finite number