Another Extremely Strange Sum

the given expression can be written as $3\sum _{ n=1 }^{ \infty }{ \frac { n }{ { 2 }^{ n } } }$

now taking only the summation part which can be expanded as $S\quad =\dfrac { 1 }{ { 2 }^{ 1 } } +\dfrac { 2 }{ { 2 }^{ 2 } } +\dfrac { 3 }{ { 2 }^{ 3 } } +\dfrac { 4 }{ { 2 }^{ 4 } } +....$ => equation (1)

$\frac { S }{ 2 } \quad = \dfrac { 1 }{ { 2 }^{ 2 } } +\dfrac { 2 }{ { 2 }^{ 3 } } +\dfrac { 3 }{ { 2 }^{ 4 } } +\dfrac { 4 }{ { 2 }^{ 5 } } +....$ => equation(2)

now subtract 2 from 1 and you will get simple infinite GP

thus summation part yields 2

so the final answer becomes $3\times 2$ = $\boxed{6}$

This sum can be rewritten as $3 * \Sigma_{n=1}^\infty n*(\frac{1}{2})^{n}$ .

Now we know that for $|x| \lt 1$ we have that $\Sigma_{n=1}^\infty x^{n} = \dfrac{x}{1 - x}$ .

If we differentiate both sides of this equation, (term by term on the LHS), with respect to $x$ , we find that

$\Sigma_{n=1}^\infty n*x^{n-1} = \dfrac{1}{(1 - x)^{2}} \Longrightarrow \Sigma_{n=1}^\infty n*x^{n} = \dfrac{x}{(1 - x)^{2}}$ .

So with $x = \frac{1}{2}$ we find that the original sum is equal to

$3 * \dfrac{\frac{1}{2}}{(1 - \frac{1}{2})^{2}} = 3 * \dfrac{\frac{1}{2}}{\frac{1}{4}} = 3 * 2 = \boxed{6}$ .