$$
\begin{array}{l}
\text { 5. Let } a_{1}=b_{1}=1, a_{n+1}=2 b_{n}+1, \\
b_{n+1}=2 a_{n}-1(n=1,2, \cdots) .
\end{array}
$$

If $d$ is a real constant such that for every positive
integer $k(k \geqslant 2)$, we have
$$
a_{2 k}+a_{2 k-1} \geqslant d\left(a_{2 k+2}+a_{2 k+1}\right),
$$

then the maximum value of $d$ is ( ).
(A) $\frac{2}{7}$
(B) $\frac{1}{4}$
(C) $\frac{1}{2}$
(D) 1
$$
5.B.

Obviously, $a_{1}=1, b_{1}=1, a_{2}=2 b_{1}+1=3$, $a_{n+1}=2 b_{n}+1=2\left(2 a_{n-1}-1\right)+1=4 a_{n-1}-1(n \geqslant 2)$.
Let $d$ be an undetermined constant, then
$$
a_{n+1}+d=4 a_{n-1}+d-1=4\left(a_{n-1}+\frac{d-1}{4}\right) \text {. }
$$

Let $d=\frac{d-1}{4}$, we get $d=-\frac{1}{3}$.
Thus, $a_{n+1}-\frac{1}{3}=4\left(a_{n-1}-\frac{1}{3}\right)$.
Therefore, $a_{2 k}-\frac{1}{3}=4\left(a_{2(k-1)}-\frac{1}{3}\right)=\cdots$
$=4^{k-1}\left(a_{2 \times 1}-\frac{1}{3}\right)=4^{k-1}\left(3-\frac{1}{3}\right)$
$=4^{k-1} \times \frac{8}{3}=\frac{2}{3} \times 4^{k}$,
i.e., $a_{2 k}=\frac{2}{3} \times 4^{k}+\frac{1}{3}(k \geqslant 1)$.
Also, $a_{2 k-1}-\frac{1}{3}=4\left(a_{2(k-1)-1}-\frac{1}{3}\right)=\cdots$
$=4^{k-1}\left(a_{2 \times 1-1}-\frac{1}{3}\right)=4^{k-1}\left(a_{1}-\frac{1}{3}\right)$
$=4^{k-1}\left(1-\frac{1}{3}\right)=\frac{2}{3} \times 4^{k-1}$,

i.e., $a_{2 k-1}=\frac{2 \times 4^{k-1}+1}{3}(k \geqslant 1)$.
(1) + (2) gives
$$
a_{2 k}+a_{2 k-1}=\frac{10}{3} \times 4^{k-1}+\frac{2}{3}(k \geqslant 1) \text {. }
$$

Substituting into the inequality in the problem, we get
$$
\frac{2}{3}\left(5 \times 4^{k-1}+1\right) \geqslant d \cdot \frac{2}{3}\left(5 \times 4^{k}+1\right) \text {, }
$$

i.e., $d \leqslant \frac{5 \times 4^{k-1}+1}{5 \times 4^{k}+1}=\frac{5 \times 4^{k}+4}{4\left(5 \times 4^{k}+1\right)}$
$$
=\frac{1}{4}\left(1+\frac{3}{5 \times 4^{k}+1}\right)(k=1,2, \cdots) .
$$

Since $\lim _{k \rightarrow+\infty} \frac{3}{5 \times 4^{k}+1}=0$, the maximum value of $d$ is $\frac{1}{4}$.