Question by FA Liên Quân Garena - Discuss with MathYouLike

FA Liên Quân Garena

08/01/2018 at 22:05

Consider the expression :

\(S=\dfrac{1}{2^0}+\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{1992}{2^{1991}}\)

Prove that : S < 4

List of answers

FA KAKALOTS 10/02/2018 at 14:23

$S = \frac{1}{2^{0}} + \frac{2}{2^{1}} + \frac{3}{2^{2}} + . . . + \frac{1992}{2^{1991}}$

$\Rightarrow 2 S = 2 + \frac{2}{1} + \frac{3}{2} + . . . + \frac{1992}{2^{1990}}$

$\Rightarrow 2 S - S = (2 + \frac{2}{1} + \frac{3}{2} + . . . + \frac{1992}{2^{1990}}) - (\frac{1}{2^{0}} + \frac{2}{2^{1}} + \frac{3}{2^{2}} + . . . + \frac{1992}{2^{1991}})$

$\Rightarrow S = 2 + \frac{1}{2^{0}} + \frac{1}{2^{1}} + \frac{1}{2^{2}} + . . . + \frac{1}{2^{1990}} - \frac{1992}{2^{1991}}$
Dao Trong Luan Coordinator 09/01/2018 at 12:29

\(S=\dfrac{1}{2^0}+\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{1992}{2^{1991}}\)

\(\Rightarrow2S=2+\dfrac{2}{1}+\dfrac{3}{2}+...+\dfrac{1992}{2^{1990}}\)

\(\Rightarrow2S-S=\left(2+\dfrac{2}{1}+\dfrac{3}{2}+...+\dfrac{1992}{2^{1990}}\right)-\left(\dfrac{1}{2^0}+\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{1992}{2^{1991}}\right)\)

\(\Rightarrow S=2+\dfrac{1}{2^0}+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{1990}}-\dfrac{1992}{2^{1991}}\)

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