tính A= 1\3+1\3^2+1\3^3+1\3^4+1\3^5+1\3^6+1\3^7+1\3^8+1\3^9+1\3^10 dấu \ là phần nha như 1\3 là 1 phần 3 03/08/2021 Bởi Rose tính A= 1\3+1\3^2+1\3^3+1\3^4+1\3^5+1\3^6+1\3^7+1\3^8+1\3^9+1\3^10 dấu \ là phần nha như 1\3 là 1 phần 3
Đáp án: `A=(1-1/3^10)/2` Giải thích các bước giải: `A=1/3+1/3^2+1/3^3+…+1/3^10``3A=1+1/3+1/3^2+1/3^3+…+1/3^9``3A-A=(1+1/3+1/3^2+1/3^3+…+1/3^9)-(1/3+1/3^2+1/3^3+…+1/3^10)``A.(3-1)=1+1/3+1/3^2+1/3^3+…+1/3^9-1/3-1/3^2-1/3^3-…-1/3^10``A2=1+(1/3-1/3)+(1/3^2-1/3^2)+…+(1/3^9-1/3^9)-1/3^10``A2=1-1/3^10``A=(1-1/3^10)/2` Bình luận
A= $\frac{1}{3}$ + $\frac{1}{3^2}$+ $\frac{1}{3^3}$+…+ $\frac{1}{3^10}$ $\frac{A}{3}$ = $\frac{1}{3^2}$+ $\frac{1}{3^3}$+…+ $\frac{1}{3^1 1}$ A -$\frac{A}{3}$= ($\frac{1}{3}$ + $\frac{1}{3^2}$+ $\frac{1}{3^3}$+…+ $\frac{1}{3^10}$)-($\frac{1}{3^2}$+ $\frac{1}{3^3}$+…+ $\frac{1}{3^1 1}$) $\frac{2}{3}$.A=$\frac{1}{3}$-$\frac{1}{3^1 1}$ A=($\frac{1}{3}$-$\frac{1}{3^1 1}$):$\frac{2}{3}$ A=($\frac{1}{3}$-$\frac{1}{3^ 1 1}$).$\frac{3}{2}$ CHUCBANHOKTOT ^^ Bình luận
Đáp án:
`A=(1-1/3^10)/2`
Giải thích các bước giải:
`A=1/3+1/3^2+1/3^3+…+1/3^10`
`3A=1+1/3+1/3^2+1/3^3+…+1/3^9`
`3A-A=(1+1/3+1/3^2+1/3^3+…+1/3^9)-(1/3+1/3^2+1/3^3+…+1/3^10)`
`A.(3-1)=1+1/3+1/3^2+1/3^3+…+1/3^9-1/3-1/3^2-1/3^3-…-1/3^10`
`A2=1+(1/3-1/3)+(1/3^2-1/3^2)+…+(1/3^9-1/3^9)-1/3^10`
`A2=1-1/3^10`
`A=(1-1/3^10)/2`
A= $\frac{1}{3}$ + $\frac{1}{3^2}$+ $\frac{1}{3^3}$+…+ $\frac{1}{3^10}$
$\frac{A}{3}$ = $\frac{1}{3^2}$+ $\frac{1}{3^3}$+…+ $\frac{1}{3^1 1}$
A -$\frac{A}{3}$= ($\frac{1}{3}$ + $\frac{1}{3^2}$+ $\frac{1}{3^3}$+…+ $\frac{1}{3^10}$)-($\frac{1}{3^2}$+ $\frac{1}{3^3}$+…+ $\frac{1}{3^1 1}$)
$\frac{2}{3}$.A=$\frac{1}{3}$-$\frac{1}{3^1 1}$
A=($\frac{1}{3}$-$\frac{1}{3^1 1}$):$\frac{2}{3}$
A=($\frac{1}{3}$-$\frac{1}{3^ 1 1}$).$\frac{3}{2}$
CHUCBANHOKTOT ^^