Question

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C=1/(-5)+1/(-5)^2+1/(-5)^3+…+1/(-5)^99

(x-3y)^2006+(y+4)^2008=0

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Answer 1

\(C=\dfrac{-1}{5}+\left(\dfrac{1}{-5}\right)^2+\left(-\dfrac{1}{5}\right)^3+...+\left(-\dfrac{1}{5}\right)^{99}\)

=>\(5\cdot C=-1+\left(-\dfrac{1}{5}\right)+\left(-\dfrac{1}{5}\right)^2+...+\left(-\dfrac{1}{5}\right)^{98}\)

=>\(5\cdot C-C=\left(-1\right)-\left(-\dfrac{1}{5}\right)^{99}\)

=>\(4C=-1+\dfrac{1}{5^{99}}=\dfrac{-5^{99}+1}{5^{99}}\)

=>\(C=\dfrac{-5^{99}+1}{4\cdot5^{99}}\)

(x-3y)^2006+(y+4)^2008=0

=>x-3y=0 và y+4=0

=>x=3y và y=-4

=>x=3*(-4)=-12 và y=-4