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 $3C=-1+\dfrac{1}{3}-\dfrac{1}{3^2}+....+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}$
$3C+C=-1-\dfrac{1}{3^{99}}$
$4C=-1-\dfrac{1}{3^{99}}$
$C=\dfrac{-1-\dfrac{1}{3^{99}}}{4}$