Let
$$a _ { n } = \int _ { 0 } ^ { 1 } x ^ { 2n } \sqrt { 1 - x ^ { 2 } } \, d x \quad ( n = 0,1,2 , \cdots )$$
We are to find the value of the limit $\lim _ { n \rightarrow \infty } \frac { a _ { n } } { a _ { n - 1 } }$.
(1) First, let us find $a _ { 0 }$ and $a _ { 1 }$. Since the area of a circle with the radius 1 is $\pi$, we see that
$$a _ { 0 } = \int _ { 0 } ^ { 1 } \sqrt { 1 - x ^ { 2 } } \, d x = \frac { \pi } { \mathbf { A } }$$
Next, the partial integral method applied to $a _ { 1 }$ gives
$$\begin{aligned}
a _ { 1 } & = \int _ { 0 } ^ { 1 } x ^ { 2 } \sqrt { 1 - x ^ { 2 } } \, d x \\
& = - \frac { \mathbf { B } } { \mathbf { C } } \left[ x \left( 1 - x ^ { 2 } \right) ^ { \frac { \mathbf { D } } { \mathbf { E } } } \right] _ { 0 } ^ { 1 } + \frac { \mathbf { F } } { \mathbf { G } } \int _ { 0 } ^ { 1 } \left( 1 - x ^ { 2 } \right) ^ { \frac { \mathbf { H } } { \mathbf{I} } } d x \\
& = \frac { \mathbf { J } } { \mathbf { K } } \left\{ \int _ { 0 } ^ { 1 } \sqrt { 1 - x ^ { 2 } } \, d x - \int _ { 0 } ^ { 1 } x ^ { \mathbf { L } } \sqrt { 1 - x ^ { 2 } } \, d x \right\}
\end{aligned}$$
Thus we have
$$a _ { 1 } = \frac { \pi } { \mathbf { M N } } .$$
(2) For $\mathbf { O } \sim \mathbf { U }$ in the following sentences, choose the correct answer from among choices (0) $\sim$ (9) below.
When the partial integral method is applied to $a _ { n }$ in the same way as for $a _ { 1 }$, we get
$$a _ { n } = \frac { \mathbf { O } } { \mathbf { P } } \left\{ \int _ { 0 } ^ { 1 } x ^ { \mathbf { Q } } \sqrt { 1 - x ^ { 2 } } \, d x - \int _ { 0 } ^ { 1 } x ^ { \mathbf { R } } \sqrt { 1 - x ^ { 2 } } \, d x \right\} \quad ( n = 1,2,3 , \cdots )$$
Hence we have
$$( \mathbf { S } ) a _ { n } = ( \mathbf { T } ) a _ { n - 1 } ,$$
and so
$$\lim _ { n \rightarrow \infty } \frac { a _ { n } } { a _ { n - 1 } } = \mathbf { U }$$
(0) 0
(1) 1
(2) 2
(3) 3
(4) 4
(5) $2 n - 2$ (6) $2 n - 1$ (7) $2 n$ (8) $2 n + 1$ (9) $2 n + 2$