S = $\frac{2}{3}$ – $\frac{2}{3^2}$ + $\frac{2}{3^3}$ -$\frac{2}{3^4}$ +…+ $\frac{2}{3^89}$ -$\frac{2}{3^90}$ 03/08/2021 Bởi Eliza S = $\frac{2}{3}$ – $\frac{2}{3^2}$ + $\frac{2}{3^3}$ -$\frac{2}{3^4}$ +…+ $\frac{2}{3^89}$ -$\frac{2}{3^90}$
Đáp án: Ta có : `S = 2/3 – 2/3^2 + 2/3^3 – 2/3^4 + … + 2/3^{89} – 2/3^{90} (1)` `=> 3S = 2 – 2/3 + 2/3^2 – 2/3^3 + … + 2/3^{88} – 2/3^{89} (2)` Lấy (1) cộng cho (2) ta được : `4S = 2 – 2/3^{90}` ` = (3^{90}.2 – 2)/3^{90}` ` => S = (3^{90}.2 – 2)/(3^{90}. 4)` Giải thích các bước giải: Bình luận
`S = 2/3 – 2/{3^2} + 2/{3^3} – 2/{3^4} + …. + 2/{3^{89}} – 2/{3^{90}}` `⇔ 3S = 2 – 2/3 + 2/{3^2} – 2/{3^3} + …. + 2/{8^{88}} – 2/{3^{89}}` `⇔ 3S + S = (2 – 2/3 + 2/{3^2} – 2/{3^3} + …. + 2/{8^{88}} – 2/{3^{89}})+( 2/3 – 2/{3^2} + 2/{3^3} – 2/{3^4} + …. + 2/{3^{89}} – 2/{3^{90}})` `⇔ 4S = 2 – 2/{3^{90}}` `⇔ 4S = {2.3^{90} – 2}/{3^{90}}` `⇔ S = \frac{2(3^{90}-1)}{3^{90}.4}` `⇔ S = \frac{3^{90}-1}{3^{90}.2}`. Bình luận
Đáp án:
Ta có :
`S = 2/3 – 2/3^2 + 2/3^3 – 2/3^4 + … + 2/3^{89} – 2/3^{90} (1)`
`=> 3S = 2 – 2/3 + 2/3^2 – 2/3^3 + … + 2/3^{88} – 2/3^{89} (2)`
Lấy (1) cộng cho (2) ta được :
`4S = 2 – 2/3^{90}`
` = (3^{90}.2 – 2)/3^{90}`
` => S = (3^{90}.2 – 2)/(3^{90}. 4)`
Giải thích các bước giải:
`S = 2/3 – 2/{3^2} + 2/{3^3} – 2/{3^4} + …. + 2/{3^{89}} – 2/{3^{90}}`
`⇔ 3S = 2 – 2/3 + 2/{3^2} – 2/{3^3} + …. + 2/{8^{88}} – 2/{3^{89}}`
`⇔ 3S + S = (2 – 2/3 + 2/{3^2} – 2/{3^3} + …. + 2/{8^{88}} – 2/{3^{89}})+( 2/3 – 2/{3^2} + 2/{3^3} – 2/{3^4} + …. + 2/{3^{89}} – 2/{3^{90}})`
`⇔ 4S = 2 – 2/{3^{90}}`
`⇔ 4S = {2.3^{90} – 2}/{3^{90}}`
`⇔ S = \frac{2(3^{90}-1)}{3^{90}.4}`
`⇔ S = \frac{3^{90}-1}{3^{90}.2}`.