cho biểu thức B = 4 phần 3 +10 phần 9 + 28 phần 27 +…..+3^98+1 phần 3^98. chứng minh B<100 27/09/2021 Bởi Iris cho biểu thức B = 4 phần 3 +10 phần 9 + 28 phần 27 +…..+3^98+1 phần 3^98. chứng minh B<100
`B=4/3+10/9+28/27+…+{3^98+1}/3^98` `⇒B={3+1}/3+{3^2+1}/3^2+{3^3+1}/3^3+…+{3^98+1}/3^98` `⇒B=(1+1/3)+(1+1/3^2)+(1+1/3^3)+…+(1+1/3^98)` `⇒B=(\underbrace{1+1+1+…+1}_{\text{98 chữ số}})+(1/3+1/3^2+1/3^3+…+1/3^98)` `⇒B=98+(1/3+1/3^2+1/3^3+…+1/3^98)` `\text{Ta đặt}` `A=1/3+1/3^2+1/3^3+…+1/3^98` `⇒3A=1+1/3+1/3^2+…+1/3^97` `⇒2A=1-1/3^98` `⇒2A<1` `⇒A<1/2<1` `⇒B<98+1<100` `⇒B<100` `(đpcm)` Bình luận
Giải thích các bước giải: Ta có:$B=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+…+\dfrac{3^{98}+1}{3^{98}}$ $\to B=\dfrac{3+1}{3}+\dfrac{3^2+1}{3^2}+\dfrac{3^3+1}{3^3}+…+\dfrac{3^{98}+1}{3^{98}}$ $\to B=(1+\dfrac13)+(1+\dfrac1{3^2})+(1+\dfrac1{3^3})+…+(1+\dfrac1{3^{98}})$ $\to B=98+\dfrac13+\dfrac1{3^2}+…+\dfrac{1}{3^{98}}$ $\to 3B=3\cdot 98+\dfrac11+\dfrac1{3}+…+\dfrac{1}{3^{97}}$ $\to 3B-B=3\cdot 98-98+\dfrac11-\dfrac1{3^{98}}$ $\to 2B=2\cdot 98+1-\dfrac1{3^{98}}$ $\to 2B<2\cdot 98+2$ $\to B<98+1$ $\to B<100$ Bình luận
`B=4/3+10/9+28/27+…+{3^98+1}/3^98`
`⇒B={3+1}/3+{3^2+1}/3^2+{3^3+1}/3^3+…+{3^98+1}/3^98`
`⇒B=(1+1/3)+(1+1/3^2)+(1+1/3^3)+…+(1+1/3^98)`
`⇒B=(\underbrace{1+1+1+…+1}_{\text{98 chữ số}})+(1/3+1/3^2+1/3^3+…+1/3^98)`
`⇒B=98+(1/3+1/3^2+1/3^3+…+1/3^98)`
`\text{Ta đặt}` `A=1/3+1/3^2+1/3^3+…+1/3^98`
`⇒3A=1+1/3+1/3^2+…+1/3^97`
`⇒2A=1-1/3^98`
`⇒2A<1`
`⇒A<1/2<1`
`⇒B<98+1<100`
`⇒B<100` `(đpcm)`
Giải thích các bước giải:
Ta có:
$B=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+…+\dfrac{3^{98}+1}{3^{98}}$
$\to B=\dfrac{3+1}{3}+\dfrac{3^2+1}{3^2}+\dfrac{3^3+1}{3^3}+…+\dfrac{3^{98}+1}{3^{98}}$
$\to B=(1+\dfrac13)+(1+\dfrac1{3^2})+(1+\dfrac1{3^3})+…+(1+\dfrac1{3^{98}})$
$\to B=98+\dfrac13+\dfrac1{3^2}+…+\dfrac{1}{3^{98}}$
$\to 3B=3\cdot 98+\dfrac11+\dfrac1{3}+…+\dfrac{1}{3^{97}}$
$\to 3B-B=3\cdot 98-98+\dfrac11-\dfrac1{3^{98}}$
$\to 2B=2\cdot 98+1-\dfrac1{3^{98}}$
$\to 2B<2\cdot 98+2$
$\to B<98+1$
$\to B<100$