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 $V[N_k]$ = $k^2V[X]$ = $k^2\dfrac{1-p}{p^2}$
On
$$\left.
\begin{align}
\operatorname{var}(\,\underbrace{X_1+\cdots+X_1}_\text{$k$ equal terms}\,)&= \operatorname{var}(kX_1) = k^2 \operatorname{var}(X_1)\\[15pt]
\operatorname{var}(\,\overbrace{X_1+\cdots+X_k}^\text{$k$ independent terms}\,)&= \operatorname{var}(X_1)+\cdots+\operatorname{var}(X_n) = k\operatorname{var}(X_1)
\end{align}
\right\}\stackrel{\text{Two}}{\text{different answers.}}
$$
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.
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 $$V(N_k) = V(X_1)+\dotsb+V(X_k) = k\cdot\frac{1-p}{p^2}.$$