From Basic Probability Theory by Ash:
Let $R$ be the number of successes in $n$ Bernoulli trials, with probability $p$ of success on a given trial. Find the conditional expectation of $R$ given that $R \geq 2$.
My thoughts: We know that the probability that the number of success is greater than 2 is $ \Sigma_{i=2}^{n} \binom{n}{i} p^i \cdot p^{n-i}$. Then the probability of each event occurring $(k=2,3,4)$ = $\binom{n}{k} p^k \cdot p^{n-k}$. So for each event we say the conditional expectation is $\frac{k \cdot \binom{n}{k} p^k \cdot p^{n-k}}{\Sigma_{i=2}^{n} \binom{n}{i} p^i \cdot p^{n-i}}$, so then the total probability will be $$\Sigma_{k=2}^{n} \left [\frac{k \cdot \binom{n}{k} p^k \cdot p^{n-k}}{\Sigma_{i=2}^{n} \binom{n}{i} p^i \cdot p^{n-i}} \right]$$
Something feels kind of murky about this, and when I write it out I always get strange answers.
What you have tried so far is correct.
$$\begin{align} \mathsf E(R\mid R\geq 2) & = \sum_{k=0}^n k\; \mathsf P(R=k\mid R\geq 2) \\[2ex] & = \frac{\sum\limits_{k=\color{blue}{2}}^n k\;\mathsf P(R=k)}{\mathsf P(R\geq 2)} \\[2ex] & = \frac{\sum\limits_{k=2}^n \binom{n}{k} k\; p^k\;(1-p)^{n-k}}{\sum\limits_{j=2}^n \binom{n}{j}\;p^j\;(1-p)^{n-j}} & \color{green}{\checkmark} \end{align}$$
Now consider though that: $$\mathsf P(R\geq 2)=1-\mathsf P(R<2)$$
Consider also that: $$\sum\limits_{k=\color{blue}{0}}^n k\; \mathsf P(R=k) = np$$
Another method By the Law of Iterated Expectation: $$\begin{align} \mathsf E(R) & = \mathsf E(R\mid R<2)\mathsf P(R< 2) + \mathsf E(R\mid R\geq 2)\mathsf P(R\geq 2) \\[2ex] \therefore \mathsf E(R\mid R\geq 2) &= \dfrac{\mathsf E(R)-\mathsf E(R\mid R<2)\mathsf P(R< 2)}{\mathsf P(R\geq 2)} \\[1ex] & = \dfrac{\mathsf E(R) - \sum_{k=0}^1 k\;\mathsf P(R=k\mid R<2)\mathsf P(R<2)}{1-\mathsf P(R<2)} \end{align}$$