Toán Tính `1/3-2/3^2+3/3^3+4/3^4+…+99/3^99-100/3^100`. 11/07/2021 By Brielle Tính `1/3-2/3^2+3/3^3+4/3^4+…+99/3^99-100/3^100`.
Đặt `A=1/3-2/3^2+3/3^3-4/3^4+…+99/3^99-100/3^100` `⇒3A=1-2/3+3/3^2-4/3^3+…+99/3^98-100/3^99` `⇒3A+A=4A=1-1/3+1/3^2-1/3^3+…+1/3^98-1/3^99-100/3^100` $⇒ -\dfrac 13.4A=-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+..+\dfrac{1}{3^{100}}+\dfrac{100}{3^{101}}$ $⇒ 4A-(-\dfrac 13.4A)=1-\dfrac{99}{3^{100}}+\dfrac{100}{3^{101}}$ $⇒ \dfrac{16A}{3}=1-\dfrac{1}{3^{100}}(99-\dfrac{100}{3})$ `⇒A=48-1/(3^99 . 16)(99-100/3)` Trả lời
Đặt `A=1/3-2/3^2+3/3^3-4/3^4+…+99/3^99-100/3^100`
`⇒3A=1-2/3+3/3^2-4/3^3+…+99/3^98-100/3^99`
`⇒3A+A=4A=1-1/3+1/3^2-1/3^3+…+1/3^98-1/3^99-100/3^100`
$⇒ -\dfrac 13.4A=-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+..+\dfrac{1}{3^{100}}+\dfrac{100}{3^{101}}$
$⇒ 4A-(-\dfrac 13.4A)=1-\dfrac{99}{3^{100}}+\dfrac{100}{3^{101}}$
$⇒ \dfrac{16A}{3}=1-\dfrac{1}{3^{100}}(99-\dfrac{100}{3})$
`⇒A=48-1/(3^99 . 16)(99-100/3)`