Question

[Đào Ngọc Lâm]

1/4²+1/4³+....+1/4¹⁰⁰

^{giúp mik với}

^{mình sắp phải nộp rùi}

Answer 1

[Nguyễn Thị Thương Hoài]

        A =          \(\dfrac{1}{4^2}\) + \(\dfrac{1}{4^3}\) + ...........+ \(\dfrac{1}{4^{100}}\)

       A =          \(\dfrac{1}{4^2}\) +  \(\dfrac{1}{4^3}\)+...+ \(\dfrac{1}{4^{99}}\)+  \(\dfrac{1}{4^{100}}\)

4 \(\times\) A =   \(\dfrac{1}{4}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{4^3}\) +...+ \(\dfrac{1}{4^{99}}\)

4A - A =   \(\dfrac{1}{4}\) - \(\dfrac{1}{4^{100}}\)

      3A =  \(\dfrac{1}{4}\) - \(\dfrac{1}{4^{100}}\)

        A = ( \(\dfrac{1}{4}\) - \(\dfrac{1}{4^{100}}\)): 3

        A =   \(\dfrac{1}{12}\) - \(\dfrac{1}{3\times4^{100}}\)

Answer 2

[Phạm Ngọc Anh]

Đặt A=1/4^2 +...+1/4^100

       4A=1/4+...+1/4^99

      4A-A=(1/4+...+1/4^99)-(1/4^2+...+1/4^100)

     3A=1/4-1/4^100

      A=(1/4-1/4^100)/3

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