4B. Let $P(x)=(2+3x)^{1994}$. Calculate the sum of all coefficients of the polynomial $P(x)$ that stand before the odd powers of $x$.

Note. The final results should be given in the shortest form, i.e., sums of a large number of addends should not appear.
Solution. We have

$$
\begin{aligned}
P(x) & =(2+3 x)^{1994} \\
& =2^{1994}+\binom{1994}{1} 2^{1993} \cdot 3 x+\ldots+\left(\begin{array}{c}
1994
\end{array}\right) 2 \cdot 3^{1993} x^{1993}+3^{1994} x^{1994}
\end{aligned}
$$

Let's denote the sum of all coefficients of the polynomial $P(x)$ by $s$, and the sum of all coefficients of the polynomial $P(x)$ that stand before even powers of $x$ by $t$.

From the expanded form of $P(x)$, it is seen that $s$ is equal to the value of $P(x)$ for $x=1$, i.e., $s=P(1)=5^{1994}$.

Let's consider $P(1)+P(-1)$. In this sum, the coefficients before even powers of $x$ are added twice (once in $P(1)$ and once in $P(-1)$), while the coefficients before odd powers of $x$ are not included at all (or rather, they cancel out in the summation because they are equal in absolute value but have different signs). Therefore:

$$
t=\frac{P(1)+P(-1)}{2}=\frac{5^{1994}+1}{2}
$$