Let $N_k$ be the number of independent trials we need to get $k$ successes when success has probability $p$. Find the mean and variance of $N_k$. 
Let $N_k$ be the number of independent trials we need to get $k$ successes when success has probability $p$. Find the mean and variance of $N_k$.

Here is my thought process so far:
Let $X =$ the number of trials to get $1$ success with probability $p$.
Then $E[N_k]$ = $kE[X]$ = $k\dfrac{1}{p}$ = $\dfrac{k}{p}$.
And $V[N_k]$ = $k^2V[X]$ = $k^2\dfrac{1-p}{p^2}$ 
 A: $$\left.
\begin{align}
\operatorname{var}(\,\underbrace{X_1+\cdots+X_1}_\text{$k$ equal terms}\,)&= \operatorname{var}(kX_1)   = k^2 \operatorname{var}(X_1)\\[15pt]
 \operatorname{var}(\,\overbrace{X_1+\cdots+X_k}^\text{$k$ independent terms}\,)&=  \operatorname{var}(X_1)+\cdots+\operatorname{var}(X_n)  = k\operatorname{var}(X_1)
\end{align}
\right\}\stackrel{\text{Two}}{\text{different answers.}}
$$
Above,
$X_1$ is the number of trials needed to get the first success;
$X_2$ is the number of trials needed after that to get the second success;
$X_3$ is the number of trials needed after that to get the second success;
and so on.  Those are independent, so the first equality about is not applicable, and the second is.
But your answer for the expected value is correct.
A: You can notice that this follows a negative binomial distribution. Then those values are well known.
Alternatively, I will add more details to your answer. 
We notice that the number of trials until the first success follows a geometric distribution $X_1\sim\text{Geom}(p)$ on $\{1,2,3,4,\dotsc\}$. Then after the first success, the number of trials until the second success also follows a geometric distribution $X_2\sim\text{Geom}(p)$ by independence. Notice that the number of trials until the second success is
$$N_2 = X_1+X_2.$$
In general, we have 
$$N_k = X_1+\dotsb+X_k$$
and so 
$$E[N_k] = E[X_1]+\dotsb+E[X_k] = k\cdot\frac{1}{p}$$
and using independence, we have
$$V(N_k) = V(X_1)+\dotsb+V(X_k) = k\cdot\frac{1-p}{p^2}.$$
