tính nhanh (1-1/1+2)*(1-1/1+2+3)*…*(1- 1/1+2+3+..+2020) 11/08/2021 Bởi Caroline tính nhanh (1-1/1+2)*(1-1/1+2+3)*…*(1- 1/1+2+3+..+2020)
Đáp án: $A=\dfrac{337}{1010}$ Giải thích các bước giải: Ta có: $I_n=1-\dfrac{1}{1+2+3+…+n}$ $\to I_n=\dfrac{1+2+3+…+n-1}{1+2+3+…+n}$ $\to I_n=\dfrac{2+3+…+n}{1+2+3+…+n}$ $\to I_n=\dfrac{\dfrac{\left(n-1\right)\left(n+2\right)}{2}}{\dfrac{n\left(n+1\right)}{2}}$ $\to I_n=\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}$ Khi đó $A=\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)….\left(1-\dfrac{1}{1+2+3+…+2020}\right)$ $\to A=\dfrac{\left(2-1\right)\left(2+2\right)}{2\left(2+1\right)}\cdot \dfrac{\left(3-1\right)\left(3+2\right)}{3\left(3+1\right)}….\dfrac{\left(2020-1\right)\left(2020+2\right)}{2020\left(2020+1\right)}$ $\to A=\dfrac{1\cdot 4}{2\cdot 3}\cdot \dfrac{2\cdot 5}{3\cdot 4}….\dfrac{2019\cdot 2022}{2020\cdot 2021}$ $\to A=\dfrac{1.2…2019}{2.3…2020}\cdot \dfrac{4.5…2022}{3.4…2021}$ $\to A=\dfrac{1}{2020}\cdot \dfrac{2022}{3}$ $\to A=\dfrac{337}{1010}$ Bình luận
Đáp án: Giải thích các bước giải: ( 1 – $\frac{1}{1 + 2}$ ) . ( 1 – $\frac{1}{1 + 2 + 3 }$ ) . ( 1 – $\frac{1}{1 + 2 + 3 + … + 2020}$ ) ⇒ ( 1 – $\frac{1}{3}$ ) . ( 1 – $\frac{1}{6}$ ) . ( 1 – $\frac{1}{(1+2020).2020 : 2 }$ ⇒ $\frac{2}{3}$ . $\frac{5}{6}$ …$\frac{2021.2020:2-1}{2021.2020:2}$ ⇒$\frac{4}{6}$ . $\frac{10}{12}$ …. $\frac{(2021.2020:2-1).2}{2021.2020}$ ⇒$\frac{1.4}{2.3}$ . $\frac{2.5}{3.4}$ ….$\frac{2019.2022}{2020.2021}$ ⇒$\frac{1.2.3.4….2019}{3.4.5…2020}$ . $\frac{4.5.6…2022}{3.4.5..2021}$ ⇒$\frac{2}{2020}$ . $\frac{2022}{3}$ = $\frac{4044}{6060}$ = Bình luận
Đáp án: $A=\dfrac{337}{1010}$
Giải thích các bước giải:
Ta có:
$I_n=1-\dfrac{1}{1+2+3+…+n}$
$\to I_n=\dfrac{1+2+3+…+n-1}{1+2+3+…+n}$
$\to I_n=\dfrac{2+3+…+n}{1+2+3+…+n}$
$\to I_n=\dfrac{\dfrac{\left(n-1\right)\left(n+2\right)}{2}}{\dfrac{n\left(n+1\right)}{2}}$
$\to I_n=\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}$
Khi đó
$A=\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)….\left(1-\dfrac{1}{1+2+3+…+2020}\right)$
$\to A=\dfrac{\left(2-1\right)\left(2+2\right)}{2\left(2+1\right)}\cdot \dfrac{\left(3-1\right)\left(3+2\right)}{3\left(3+1\right)}….\dfrac{\left(2020-1\right)\left(2020+2\right)}{2020\left(2020+1\right)}$
$\to A=\dfrac{1\cdot 4}{2\cdot 3}\cdot \dfrac{2\cdot 5}{3\cdot 4}….\dfrac{2019\cdot 2022}{2020\cdot 2021}$
$\to A=\dfrac{1.2…2019}{2.3…2020}\cdot \dfrac{4.5…2022}{3.4…2021}$
$\to A=\dfrac{1}{2020}\cdot \dfrac{2022}{3}$
$\to A=\dfrac{337}{1010}$
Đáp án:
Giải thích các bước giải:
( 1 – $\frac{1}{1 + 2}$ ) . ( 1 – $\frac{1}{1 + 2 + 3 }$ ) . ( 1 – $\frac{1}{1 + 2 + 3 + … + 2020}$ )
⇒ ( 1 – $\frac{1}{3}$ ) . ( 1 – $\frac{1}{6}$ ) . ( 1 – $\frac{1}{(1+2020).2020 : 2 }$
⇒ $\frac{2}{3}$ . $\frac{5}{6}$ …$\frac{2021.2020:2-1}{2021.2020:2}$
⇒$\frac{4}{6}$ . $\frac{10}{12}$ …. $\frac{(2021.2020:2-1).2}{2021.2020}$
⇒$\frac{1.4}{2.3}$ . $\frac{2.5}{3.4}$ ….$\frac{2019.2022}{2020.2021}$
⇒$\frac{1.2.3.4….2019}{3.4.5…2020}$ . $\frac{4.5.6…2022}{3.4.5..2021}$
⇒$\frac{2}{2020}$ . $\frac{2022}{3}$ = $\frac{4044}{6060}$ =