B=2^2/3 . 3^2/8 . 4^2/15. … . 100^2/999 04/09/2021 Bởi Skylar B=2^2/3 . 3^2/8 . 4^2/15. … . 100^2/999
Giải thích các bước giải: Ta có: $B=\dfrac{2^2}{3}.\dfrac{3^2}{8}.\dfrac{4^2}{15}….\dfrac{100^2}{9999}$ $\to B=\dfrac{2^2}{2^2-1}.\dfrac{3^2}{3^2-1}.\dfrac{4^2}{4^2-1}….\dfrac{100^2}{100^2-1}$ $\to B=\dfrac{2^2}{(2-1)(2+1)}.\dfrac{3^2}{(3-1)(3+1)}.\dfrac{4^2}{(4-1)(4+1)}….\dfrac{100^2}{(100-1)(100+1)}$ $\to B=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}….\dfrac{100^2}{99.101}$ $\to B=\dfrac{2.3.4…100}{1.2.3…99}.\dfrac{2.3.4….100}{3.4.5…101}$ $\to B=100.\dfrac{2}{101}$ $\to B=\dfrac{200}{101}$ Bình luận
Giải thích các bước giải:
Ta có:
$B=\dfrac{2^2}{3}.\dfrac{3^2}{8}.\dfrac{4^2}{15}….\dfrac{100^2}{9999}$
$\to B=\dfrac{2^2}{2^2-1}.\dfrac{3^2}{3^2-1}.\dfrac{4^2}{4^2-1}….\dfrac{100^2}{100^2-1}$
$\to B=\dfrac{2^2}{(2-1)(2+1)}.\dfrac{3^2}{(3-1)(3+1)}.\dfrac{4^2}{(4-1)(4+1)}….\dfrac{100^2}{(100-1)(100+1)}$
$\to B=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}….\dfrac{100^2}{99.101}$
$\to B=\dfrac{2.3.4…100}{1.2.3…99}.\dfrac{2.3.4….100}{3.4.5…101}$
$\to B=100.\dfrac{2}{101}$
$\to B=\dfrac{200}{101}$