問題 2.25

SICP 
(define list1 (list 1 3 (list 5 7) 9))

;list1
;-> (1 3 (5 7) 9)

;(car (cdr (car (cdr (cdr list1)))))
;-> 7

(define list2 (list (list 7)))

;list2
;-> ((7))

;(car (car list2))
;-> 7

(define list3 (list 1 (list 2 (list 3 (list 4 (list 5 (list 6 7)))))))

;list3
;-> (1 (2 (3 (4 (5 (6 7))))))

;(car (cdr (car (cdr (car (cdr (car (cdr (car (cdr (car (cdr list3))))))))))))
;-> 7

;ok