2024年考研数学三第4题

答案: A

解析:

$\ln (2+x)=\ln \left(1+\displaystyle\frac{1}{2} x\right)=\ln 2+\ln \left(1+\displaystyle\frac{1}{2} x\right)=\ln 2+\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1} \displaystyle\frac{\left(\displaystyle\frac{1}{2} x\right)^{n}}{n}$
所以，$a_{n}=\left{\begin{array}{l}\ln 2, n=0 \ (-1)^{n-1} \displaystyle\frac{1}{n 2^{2}}, n>0\end{array}\right.$
当 $\mathrm{n}>0$ 时，$\quad a_{2 n}=-\displaystyle\frac{1}{2 n \cdot 2^{2 n}}$
所以 $\displaystyle\sum_{n=0}^{\infty} n a_{2 n}=\displaystyle\sum_{n=1}^{\infty} n a_{2 n}=\displaystyle\sum_{n=1}^{\infty} n \cdot\left(-\displaystyle\frac{1}{2 n \cdot 2^{2 n}}\right)=-\displaystyle\sum_{n=1}^{\infty} \displaystyle\frac{1}{2^{2 n+1}}=-\displaystyle\frac{\left(\displaystyle\frac{1}{2}\right)^{3}}{1-\displaystyle\frac{1}{4}}=-\displaystyle\frac{1}{6}$
故选 A。
方法 2：$[\ln (2+x)]^{\prime}=\displaystyle\frac{1}{2+x}=\displaystyle\frac{1}{2\left(1+\displaystyle\frac{x}{2}\right)}=\displaystyle\frac{1}{2} \displaystyle\sum_{n=0}^{\infty}(-1)^{n}\left(\displaystyle\frac{x}{2}\right)^{n}$ ，
$\ln (2+x)=\displaystyle\sum_{n=0}^{\infty}(-1)^{n} \displaystyle\frac{\left(\displaystyle\frac{1}{2} x\right)^{n+1}}{n+1}+C=\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1} \displaystyle\frac{\left(\displaystyle\frac{1}{2} x\right)^{n}}{n}+C 。$
$\mathrm{S}(0)=\mathrm{C}=\ln (2+0)=\ln 2$

$$
a_{n}=\left{\begin{array}{l}
\ln 2, n=0 \
(-1)^{n-1} \frac{1}{n 2^{2}}, n>0
\end{array}\right.
$$

所以 $\displaystyle\sum_{n=0}^{\infty} n a_{2 n}=\displaystyle\sum_{n=1}^{\infty} n a_{2 n}=\displaystyle\sum_{n=1}^{\infty} n \cdot\left(-\displaystyle\frac{1}{2 n \cdot 2^{2 n}}\right)=-\displaystyle\sum_{n=1}^{\infty} \displaystyle\frac{1}{2^{2 n+1}}=-\displaystyle\frac{\left(\displaystyle\frac{1}{2}\right)^{3}}{1-\displaystyle\frac{1}{4}}=-\displaystyle\frac{1}{6}$ 。