Giúp mình với ạ! Càng nhanh càng tốt ạ! V=4×5^100x(1/5+1/5^2+1/5^3+…+1/5^100)+1 14/09/2021 Bởi Josie Giúp mình với ạ! Càng nhanh càng tốt ạ! V=4×5^100x(1/5+1/5^2+1/5^3+…+1/5^100)+1
`V=4.5^100.(1/5+1/5^2+1/5^3+….+1/5^100)+1` `V=4. (5^99+5^98+5^97+….+1)+1` Đặt `1+5+….+5^99=X->V=4X+1` ta có: `5X=5+5^2+…..+5^99+5^100` `5X-X=(5+5^2+…+5^100)-(1+5+….+5^99)` `4X=5^100-1` `X=(5^100-1)/4` `-> V=4. (5^100-1)/4+1` `V=5^100-1+1` `V=5^100` Vậy `V=5^100` Bình luận
Tham khảo Đặt `A=\frac{1}{5}+\frac{1}{5^2}+….+\frac{1}{5^{100}}` `⇒5A=1+\frac{1}{5}+\frac{1}{5^{99}}` `⇒5A-A=1+\frac{1}{5}+\frac{1}{5^{99}}-(\frac{1}{5}+\frac{1}{5^2}+….+\frac{1}{5^{100}})` `⇒4A=1-\frac{1}{5^{100}}` $⇒A=\dfrac{1-\dfrac{1}{5^{100}}}{4}$ Do đó `V=4.5^{100}×(`$\dfrac{1-\dfrac{1}{5^{100}}}{4}$`)+1` $⇒V=5^{100}.\dfrac{5^{100}-1}{5^{100}}+1$ `⇒V=5^{100}-1+1` `⇒V=5^{100}` `\text{©CBT}` Bình luận
`V=4.5^100.(1/5+1/5^2+1/5^3+….+1/5^100)+1`
`V=4. (5^99+5^98+5^97+….+1)+1`
Đặt `1+5+….+5^99=X->V=4X+1`
ta có: `5X=5+5^2+…..+5^99+5^100`
`5X-X=(5+5^2+…+5^100)-(1+5+….+5^99)`
`4X=5^100-1`
`X=(5^100-1)/4`
`-> V=4. (5^100-1)/4+1`
`V=5^100-1+1`
`V=5^100`
Vậy `V=5^100`
Tham khảo
Đặt `A=\frac{1}{5}+\frac{1}{5^2}+….+\frac{1}{5^{100}}`
`⇒5A=1+\frac{1}{5}+\frac{1}{5^{99}}`
`⇒5A-A=1+\frac{1}{5}+\frac{1}{5^{99}}-(\frac{1}{5}+\frac{1}{5^2}+….+\frac{1}{5^{100}})`
`⇒4A=1-\frac{1}{5^{100}}`
$⇒A=\dfrac{1-\dfrac{1}{5^{100}}}{4}$
Do đó `V=4.5^{100}×(`$\dfrac{1-\dfrac{1}{5^{100}}}{4}$`)+1`
$⇒V=5^{100}.\dfrac{5^{100}-1}{5^{100}}+1$
`⇒V=5^{100}-1+1`
`⇒V=5^{100}`
`\text{©CBT}`