`A= 1/(3^3) + 1/(3^5) + 1/(3^7)+….+ 1/(3^101)` Help me.Chiều nộp rùi 04/12/2021 Bởi Kaylee `A= 1/(3^3) + 1/(3^5) + 1/(3^7)+….+ 1/(3^101)` Help me.Chiều nộp rùi
$A=\frac{1}{3^3}+$ $\frac{1}{3^5}+$ $\frac{1}{3^7}+…+$ $\frac{1}{3^{101}}$ $⇒3^2A=\frac{1}{3}+$ $\frac{1}{3^3}+$ $\frac{1}{3^5}+…+$ $\frac{1}{3^{99}}$ $⇒3^2A-A=\frac{1}{3}+($ $\frac{1}{3^2}-$ $\frac{1}{3^2})+…-$ $\frac{1}{3^{101}}$ $⇒8A=\frac{1}{2}-$ $\frac{1}{3^{101}}$ $⇒A=\frac{1}{16}-$ $\frac{1}{8.3^{101}}$ $⇒A=\frac{3^{101}-2}{16.3^{101}}$ Bình luận
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$A=\frac{1}{3^3}+$ $\frac{1}{3^5}+$ $\frac{1}{3^7}+…+$ $\frac{1}{3^{101}}$
$⇒3^2A=\frac{1}{3}+$ $\frac{1}{3^3}+$ $\frac{1}{3^5}+…+$ $\frac{1}{3^{99}}$
$⇒3^2A-A=\frac{1}{3}+($ $\frac{1}{3^2}-$ $\frac{1}{3^2})+…-$ $\frac{1}{3^{101}}$
$⇒8A=\frac{1}{2}-$ $\frac{1}{3^{101}}$
$⇒A=\frac{1}{16}-$ $\frac{1}{8.3^{101}}$
$⇒A=\frac{3^{101}-2}{16.3^{101}}$