Adding I

$\sum \limits_{n=0} ^ {\infty}{{r} ^ {n}} = \frac {1} {1 -r}, -1 \leq r < 1$

Deriving both sides with respect to $r$ .

$\sum \limits_{n=0} ^ {\infty}{n{r} ^ {n-1}} = \frac {1} {{(1 -r)} ^ {2}}, -1 \leq r < 1$

$\sum \limits_{n=0} ^ {\infty}{n{r} ^ {n}} = \frac {r} {{(1 -r)} ^ {2}}, -1 \leq r < 1$

Deriving again

$\sum \limits_{n=0} ^ {\infty}{ {n} ^ {2} {r} ^ {n-1}} = \frac {1+r} {{(1 -r)} ^ {3}}, -1 \leq r < 1$

$\sum \limits_{n=0} ^ {\infty}{ {n} ^ {2} {r} ^ {n}} = \frac {r+{r} ^ {2}} {{(1 -r)} ^ {3}}, -1 \leq r < 1$

Plugging in $r = \frac{1}{2}$ yields $6$ .

$\sum _{ n=0 }^{ \infty }{ \frac { n^{ 2 } }{ 2^{ n } } } =\sum _{ n=1 }^{ \infty }{ \frac { n^{ 2 } }{ 2^{ n } } } =\sum _{ n=0 }^{ \infty }{ \frac { (n+1)^{ 2 } }{ 2^{ n+1 } } } =\sum _{ n=0 }^{ \infty }{ \frac { n^{ 2 }+2n+1 }{ 2^{ n+1 } } } =\frac { 1 }{ 2 } \sum _{ n=0 }^{ \infty }{ \frac { n^{ 2 } }{ 2^{ n } } } +\sum _{ n=0 }^{ \infty }{ \frac { n }{ 2^{ n } } } +\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ 2^{ n } } }$ Then, $\frac { 1 }{ 2 } \sum _{ n=0 }^{ \infty }{ \frac { n^{ 2 } }{ 2^{ n } } } =\sum _{ n=0 }^{ \infty }{ \frac { n }{ 2^{ n } } } +\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ 2^{ n } } } =2+1=3$ Finally $\sum _{ n=0 }^{ \infty }{ \frac { n^{ 2 } }{ 2^{ n } } } =6$