jmc

Let $$S=\sum _{k=1}^{\infty }\frac{k^2}{2^k}=\frac{1^2}{2}+\frac{2^2}{2^2}+\frac{3^2}{2^3}+\frac{4^2}{2^4}+\cdots .$$Then $$2S=1+\frac{2^2}{2}+\frac{3^2}{2^2}+\frac{4^2}{2^3}+\frac{5^2}{2^4}+\cdots .$$Subtracting these equations, we get $$S=1+\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3}+\frac{9}{2^4}+\cdots .$$Then $$2S=2+3+\frac{5}{2}+\frac{7}{2^2}+\frac{9}{2^3}+\frac{11}{2^4}+\cdots .$$Subtracting these equations, we get $$S=4+\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+\frac{2}{2^4}+\cdots =4+\frac{1}{1-1/2}=\boxed{6}.$$