In the book I am reading (complexity and cryptography by Talbot and Welsh, chapter 4), there is a theorem on $\textbf{BPP}$ where I don't understand a few steps of its proof, it's totally independent of the concept I am reading, I don't understand the mathematical steps (tagged $(1)$).
Proof:Suppose $x \in L$ , We need to show that for a suitable choice of $t$, the probability that M rejects $x$ is at most $2^{-p(n)}$ Now M rejects $x$ iff N accepts $x$ at most $t$ times during the $2t + 1$ computations.Hence: $$Pr[\text{M rejects $x$}] \le \sum_{k=0}^t \begin{pmatrix} 2t+1 \\ k \end{pmatrix} (\frac {3}{4})^k (\frac {1}{4})^{2t+1-k} \tag{1} \\\le \frac {(t+1)}{4} \begin{pmatrix} 2t+1 \\ t \end{pmatrix} (\frac {3}{16})^{t} \\ \le 2^{2t+1} \frac {(t+1)}{4} (\frac {3}{16})^{t} \le(t+1) (\frac {3}{4})^{t}$$
It follows from this that
$$\begin{align*} \Bbb P[\text{M rejects }x]&=\Bbb P[\text{N accepts }x\text{ at most }t\text{ times}]\\ &=\sum_{k=0}^t\Bbb P[\text{N accepts }x\text{ exactly }k\text{ times}] \end{align*}$$
Suppose that the probability that $N$ accepts $x$ is $\frac34$. Let $K$ be a particular set of $k$ of the $2t+1$ trials; then the probability that $N$ accepts $x$ on all of the trials in $K$ and does not accept $x$ on any of the remaining trials is
$$\left(\frac34\right)^k\left(\frac14\right)^{2t+1-k}\;.$$
There are $\binom{2t+1}k$ possible sets of $k$ trials, so the probability that $N$ accepts $x$ on exactly $k$ trials is
$$\binom{2t+1}k\left(\frac34\right)^k\left(\frac14\right)^{2t+1-k}\;,$$
and
$$\sum_{k=0}^t\binom{2t+1}k\left(\frac34\right)^k\left(\frac14\right)^{2t+1-k}$$
is the probability that $N$ accepts $x$ at most $t$ times.
Without more context I’m not sure why we have only the inequality
$$\Bbb P[\text{M rejects }x]\le\sum_{k=0}^t\binom{2t+1}k\left(\frac34\right)^k\left(\frac14\right)^{2t+1-k}$$
instead of equality, but this should at least point you in the right direction.