Tính (1+1/3).(1+1/8).(1+1/15)……………(1+1/2409) 03/11/2021 Bởi Jade Tính (1+1/3).(1+1/8).(1+1/15)……………(1+1/2409)
Đặt $Q = \bigg(1+\dfrac{1}{3}\bigg).\bigg(1+\dfrac{1}{8}\bigg)….\bigg(1+\dfrac{1}{2400}\bigg)$ $ = \dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}….\dfrac{2401}{2400}$ $ = \dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}…..\dfrac{49.49}{48.50}$ $ = \dfrac{(2.3.4…49).(2.3.4…49)}{(1.2.3…48).(3.4.5…50)}$ $ = \dfrac{49.2}{50} = \dfrac{49}{25}$ Bình luận
Đáp án: Giải thích các bước giải: Đặt A=(1+13).(1+18)....(1+12400)A=(1+13).(1+18)….(1+12400) =43.98.1615....24012400=43.98.1615….24012400 =2.21.3.3.32.4.....49.4948.50=2.21.3.3.32.4…..49.4948.50 =(2.3.4…49).(2.3.4…49)(1.2.3…48).(3.4.5…50)=(2.3.4…49).(2.3.4…49)(1.2.3…48).(3.4.5…50) =49.250=4925 anh Bình luận
Đặt $Q = \bigg(1+\dfrac{1}{3}\bigg).\bigg(1+\dfrac{1}{8}\bigg)….\bigg(1+\dfrac{1}{2400}\bigg)$
$ = \dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}….\dfrac{2401}{2400}$
$ = \dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}…..\dfrac{49.49}{48.50}$
$ = \dfrac{(2.3.4…49).(2.3.4…49)}{(1.2.3…48).(3.4.5…50)}$
$ = \dfrac{49.2}{50} = \dfrac{49}{25}$
Đáp án:
Giải thích các bước giải:
Đặt A=(1+13).(1+18)....(1+12400)A=(1+13).(1+18)….(1+12400)
=43.98.1615....24012400=43.98.1615….24012400
=2.21.3.3.32.4.....49.4948.50=2.21.3.3.32.4…..49.4948.50
=(2.3.4…49).(2.3.4…49)(1.2.3…48).(3.4.5…50)=(2.3.4…49).(2.3.4…49)(1.2.3…48).(3.4.5…50)
=49.250=4925
anh