4.1.10 problem 10

The probability that the game lasts $n$ rounds is $1∕2^n$.

Thus, if the winnings for $n$ rounds is $n$, we must compute ${\sum }_{i=1}^{\infty }(i/2^i)$.

We know that ${\sum }_{i=1}^{\infty }\left(x^i\right)=\frac{x}{1-x}$. Deriving both sides with respect to x gives ${\sum }_{i=1}^{\infty }\left(i{x}^{i-1}\right)=\frac{1}{{(1-x)}^2}$. Multiplying by both sides gives ${\sum }_{i=1}^{\infty }\left(i{x}^i\right)=\frac{x}{{(1-x)}^2}$. Plugging in $x=1∕2$ gives the answer $2$.

For the second part of the problem we need to find ${\sum }_{i=1}^{\infty }(i^2/2^i)$.

We know ${\sum }_{i=1}^{\infty }\left(i{x}^i\right)=\frac{x}{{(1-x)}^2}$. Deriving both sides with respect to x again, using the quotient rule, gives ${\sum }_{i=1}^{\infty }\left(i^2{x}^{i-1}\right)=\frac{1+x}{{(1-x)}^3}$. Multiplying both sides by $x$ gives ${\sum }_{i=1}^{\infty }\left(i^2{x}^i\right)=\frac{x+x^2}{{(1-x)}^3}$. Plugging in $x=1∕2$ gives the answer of $6$.