Question

[nguyễn đình long]

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Answer 1

[Akai Haruma]

Lời giải:

$S-1=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+...+\frac{1}{2^{99}}+\frac{1}{2^{101}}$

$2^2(S-1)=2+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+...+\frac{1}{2^{97}}+\frac{1}{2^{99}}$
$\Rightarrow 2^2(S-1)-(S-1)=2-\frac{1}{2^{101}}$

$\Rightarrow 3(S-1)=2-\frac{1}{2^{101}}$

$\Rightarrow S=\frac{1}{3}(2-\frac{1}{2^{101}})+1$

Answer 2

[Nguyễn Thị Thương Hoài]

      S =        1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^3}\) + \(\dfrac{1}{2^5}\)+ \(\dfrac{1}{2^7}\)+...+\(\dfrac{1}{2^{99}}\)+ \(\dfrac{1}{2^{101}}\)

     4S = 4 + 2 + \(\dfrac{1}{2}\) +   \(\dfrac{1}{2^3}\) + \(\dfrac{1}{2^5}\) + \(\dfrac{1}{2^7}\)+...+\(\dfrac{1}{2^{99}}\)

4S - S = 5 - \(\dfrac{1}{2^{101}}\)

    3S  = 5 - \(\dfrac{1}{2^{101}}\)

      S  = \(\dfrac{5}{3}\) - \(\dfrac{1}{3.2^{101}}\)