$\frac{1}{3}$ +$\frac{2}{ 3^{2} }$ +$\frac{3}{ 3^{3} }$ +…+$\frac{100}{ 3^{100} }$ 18/08/2021 Bởi Charlie $\frac{1}{3}$ +$\frac{2}{ 3^{2} }$ +$\frac{3}{ 3^{3} }$ +…+$\frac{100}{ 3^{100} }$
Đặt A = `1/3+2/3^2+3/3^3 +…+100/3^100` `⇒3A= 1+2/3+3/3^2 +…+100/3^99` `⇒3A-A=2A=1+1/3+1/3^2+….+1/3^99-1/3^100` `⇒A=(1+1/3+1/3^2+….+1/3^99-1/3^100)/2` Đặt `1+1/3+1/3^2+….+1/3^99=B` `⇒3B= 3+1+1/3+….+1/3^98` `⇒3B-B=2B=3-1/3^99` `⇒B=(3-1/3^99)/2` `⇒A=((3-1/3^99)/2-1/3^100)/2` Bình luận
Đặt A = `1/3+2/3^2+3/3^3 +…+100/3^100`
`⇒3A= 1+2/3+3/3^2 +…+100/3^99`
`⇒3A-A=2A=1+1/3+1/3^2+….+1/3^99-1/3^100`
`⇒A=(1+1/3+1/3^2+….+1/3^99-1/3^100)/2`
Đặt `1+1/3+1/3^2+….+1/3^99=B`
`⇒3B= 3+1+1/3+….+1/3^98`
`⇒3B-B=2B=3-1/3^99`
`⇒B=(3-1/3^99)/2`
`⇒A=((3-1/3^99)/2-1/3^100)/2`