Tính C = -$\frac{1}{3}$ + $\frac{1}{3}$ -$\frac{1}{3}$ + $\frac{1}{3^2}$ – $\frac{1}{3^3}$ + … + $\frac{1}{3^98}$ – $\frac{1}{3^99}$
Tính C = -$\frac{1}{3}$ + $\frac{1}{3}$ -$\frac{1}{3}$ + $\frac{1}{3^2}$ – $\frac{1}{3^3}$ + … + $\frac{1}{3^98}$ – $\frac{1}{3^99}$
`C = -1/3 + 1/3^2 – 1/3^3 + …..+ 1/3^98 – 1/3^99`
`3C = -1 + 1/3 – 1/3^2 + 1/3^3 – …..- 1/3^98 +1/3^99`
`3C+C = (-1 + 1/3 – 1/3^2 + 1/3^3 – …..- 1/3^98 +1/3^99) + ( -1/3 + 1/3^2 – 1/3^3 + …..+ 1/3^98 – 1/3^99)`
`4C = -1 – 1/3^99`
`C = (-1 – 1/3^99)/4`
Đáp án+Giải thích các bước giải:
$C=-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{3}+….+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}$
$C=-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+….+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}$
Đặt $3C=-1+\dfrac{1}{3}-\dfrac{1}{3^2}+….+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}$
$3C+C=-1-\dfrac{1}{3^{99}}$
$4C=-1-\dfrac{1}{3^{99}}$
$C=\dfrac{-1-\dfrac{1}{3^{99}}}{4}$