Analysis the following code:

int RunningSum$($int n$)$

$\{$

if $($ n$==1)$

return $1$;

else

return $($n $+$ RunningSum$($n$-1))$;

$\}$

During execution of the function call RunningSum$(4),$ RunningSum makes a function call to itself, with an argument of $3($i$.$e$.,$ RunningSum$(3)).$ However, before RunningSum$(3)$ ends, it makes a call to RunningSum$(2).$ And before RunningSum$(2)$ ends, it makes a call to RunningSum$(1).$ RunningSum$(1),$ however, makes no additional recursive calls and returns the value $1$ to RunningSum$(2),$ which enables RunningSum$(2)$ to end and return the value $2+1$ back to RunningSum$(3).$ This enables RunningSum$(3)$ to end and pass a value of $3+2+1$ to RunningSum$(4).$ The image pictorially shows how the execution of RunningSum$(4)$ proceeds.

Now, answer the following:

$1.$ How many calls to RunningSum are made for the call RunningSum$(10)?$ Explain your answer.

$2.$ How about for the call RunningSum$($n$)?$ Give your answer in terms of n$.$ Explain your answer.