Đáp án:
$⇒D=(1-\dfrac{1}{3^{204}})÷8$ hay $\dfrac{1-\dfrac{1}{3^{204}}}{8}$
Giải thích các bước giải:
$D=\dfrac{1}{3²}-\dfrac{1}{3^4}+..+\dfrac{1}{3^{202}}- \dfrac{1}{3^{204}}$
Cách làm:Nhân vế trái,vế phải với $3²$
$⇒3²D=3²×(\dfrac{1}{3²}-\dfrac{1}{3^4}+..+\dfrac{1}{3^{202}}- \dfrac{1}{3^{204}})$
$⇒9D=1-\dfrac{1}{3²}+..+\dfrac{1}{3^{200}} \dfrac{1}{3^{202}}$
$⇒9D-D=(1-\dfrac{1}{3²}+..+\dfrac{1}{3^{200}} \dfrac{1}{3^{202}})-\dfrac{1}{3²}-\dfrac{1}{3^4}+..+\dfrac{1}{3^{202}}- \dfrac{1}{3^{204}} $
$⇒8D=1-\dfrac{1}{3^{204}} $
$⇒D=(1-\dfrac{1}{3^{204}})÷8$ hay $\dfrac{1-\dfrac{1}{3^{204}}}{8}$