Toán 2n:[1+(1\1+2)+(1\1+2+3)+…+(1\1+2+3+4)+…+(1+1+2+3+4+…+n)]=2020 tìm n 09/09/2021 By Valerie 2n:[1+(1\1+2)+(1\1+2+3)+…+(1\1+2+3+4)+…+(1+1+2+3+4+…+n)]=2020 tìm n
Giải thích các bước giải: Ta có $1+2+3+…+n=\dfrac{n(n+1)}{2}$ $\to \dfrac1{1+2+3+..+n}=\dfrac{2}{n(n+1)}=2\cdot \dfrac{ n+1-n}{n(n+1)}=2\cdot (\dfrac1n-\dfrac1{n+1})$ Áp dụng ta có: $P=\dfrac1{1+2}+\dfrac1{1+2+3}+…+\dfrac{1}{1+2+3+…+n}$ $P=2\cdot (\dfrac12-\dfrac13)+2\cdot (\dfrac13-\dfrac14)+…+2\cdot( \dfrac1{n-1}-\dfrac1{n})$ $\to P=2\cdot (\dfrac12-\dfrac13+\dfrac13-\dfrac14+…+\dfrac1{n-1}-\dfrac1{n})$ $\to P=2\cdot (\dfrac12-\dfrac1{n})$ $\to P=1-\dfrac{2}{n}$ $\to P=\dfrac{n-2}{n}$ $\to 1+\dfrac1{1+2}+…+\dfrac1{1+2+3+…+n}=1+\dfrac{n-2}{n}=\dfrac{2n-2}{n}$ $\to \dfrac{2n}{1+\dfrac1{1+2}+…+\dfrac1{1+2+3+…+n}}=\dfrac{2n}{\dfrac{2n-2}{n}}=\dfrac{n^2}{n-1}$ $\to \dfrac{n^2}{n-1}=2020$ $\to n^2=2020(n-1)$ $\to n^2=2020n-2020$ $\to n^2-2020n+2020=0$ $\to $Không có nghiệm nguyên $\to$Không tồn tại $n$ thỏa mãn đề Trả lời
Giải thích các bước giải:
Ta có $1+2+3+…+n=\dfrac{n(n+1)}{2}$
$\to \dfrac1{1+2+3+..+n}=\dfrac{2}{n(n+1)}=2\cdot \dfrac{ n+1-n}{n(n+1)}=2\cdot (\dfrac1n-\dfrac1{n+1})$
Áp dụng ta có:
$P=\dfrac1{1+2}+\dfrac1{1+2+3}+…+\dfrac{1}{1+2+3+…+n}$
$P=2\cdot (\dfrac12-\dfrac13)+2\cdot (\dfrac13-\dfrac14)+…+2\cdot( \dfrac1{n-1}-\dfrac1{n})$
$\to P=2\cdot (\dfrac12-\dfrac13+\dfrac13-\dfrac14+…+\dfrac1{n-1}-\dfrac1{n})$
$\to P=2\cdot (\dfrac12-\dfrac1{n})$
$\to P=1-\dfrac{2}{n}$
$\to P=\dfrac{n-2}{n}$
$\to 1+\dfrac1{1+2}+…+\dfrac1{1+2+3+…+n}=1+\dfrac{n-2}{n}=\dfrac{2n-2}{n}$
$\to \dfrac{2n}{1+\dfrac1{1+2}+…+\dfrac1{1+2+3+…+n}}=\dfrac{2n}{\dfrac{2n-2}{n}}=\dfrac{n^2}{n-1}$
$\to \dfrac{n^2}{n-1}=2020$
$\to n^2=2020(n-1)$
$\to n^2=2020n-2020$
$\to n^2-2020n+2020=0$
$\to $Không có nghiệm nguyên
$\to$Không tồn tại $n$ thỏa mãn đề