11.4 If $(1.0025)^{10}$ is calculated to five decimal places, then the digit in the fifth decimal place is
(A) 0 .
(B) 1 .
(C) 2 .
(D) 5 .
(E) 8 .
(22nd American High School Mathematics Examination, 1971)
［Solution］By the binomial expansion, we have
$$
\begin{aligned}
(1.0025)^{10}= & (1+0.0025)^{10} \\
= & 1+10(0.0025)+45(0.0025)^{2}+120(0.0025)^{3} \\
& +210(0.0025)^{4}+\cdots \\
= & 1.025+0.00028125+0.000001875+R \\
= & 1.025283125+R .
\end{aligned}
$$

where $R$ is the sum of terms smaller than $10^{-8}$, i.e., $R=\sum_{k=4}^{10} C_{10}^{k} b^{k}$, where $b=0.0025$.
Since $0<b<1$, and $\sum_{k=0}^{10} C_{10}^{k}=(1+1)^{10}=2^{10}$, we have $R=\sum_{k=4}^{10} C_{10}^{k} b^{k}<b^{4} \sum_{k=4}^{10} C_{10}^{k}<b^{4} \sum_{k=0}^{10} C_{10}^{k}=b^{4} \cdot 2^{10}$.
Thus, $R<\frac{2^{10}}{2^{8} \cdot 10^{8}}=\frac{2^{2}}{10^{8}}=0.00000004$.
$\therefore \quad(1.0025)^{10} \approx 1.02528$ (accurate to 5 decimal places). Hence, the answer is $(E)$.