My problems #10

Algebra Level 3

The sum of the infinite series

$\dfrac{\text{log}2}{3}+\dfrac{\text{log}4}{9}+\dfrac{\text{log}8}{27}+\ldots$

can be expressed as $\displaystyle\ \dfrac{a\times{\text{log}\ b}}{c}.$ Find $a+b+c$

The answer is 9.

1 solution

Saurav Pal
Apr 2, 2015

$Let\quad \frac { log2 }{ 3 } +\frac { log4 }{ 9 } +\frac { log8 }{ 27 } +.\quad .\quad .=\quad x.\\ x\quad =\quad \frac { log2 }{ 3 } +\frac { 2log2 }{ { 3 }^{ 2 } } +\frac { 3log2 }{ { 3 }^{ 3 } } +.\quad .\quad .\\ \frac { x }{ 3 } \quad =\quad \quad \quad \quad \quad \quad \frac { log2 }{ { 3 }^{ 2 } } +\frac { 2log2 }{ { 3 }^{ 3 } } +.\quad .\quad .\\ \frac { 2x }{ 3 } \quad =\quad \frac { log2 }{ 3 } +\frac { log2 }{ { 3 }^{ 2 } } +\frac { log2 }{ { 3 }^{ 3 } } +.\quad .\quad .\\ 2x\quad =\quad log2\quad (1+\frac { 1 }{ 3 } +\frac { 1 }{ { 3 }^{ 2 } } +.\quad .\quad .)\\ 2x\quad =\quad \frac { 3 }{ 2 } log2\quad \Rightarrow \quad x\quad =\quad \boxed { \frac { 3\quad log2 }{ 4 } } .\\ \therefore \quad a+b+c\quad =\quad 3+2+4\quad =\quad \boxed { 9 } .$

Could you please explain how do you get the fourth line (the 2x/3 part)?

谦艺伍 - 4 years, 9 months ago

I believe he subtracted the x/3 series from the series for x.

Tristan Goodman - 1 year, 4 months ago

Oh I see. Thanks.

谦艺伍 - 1 year, 2 months ago

My problems #10

The answer is 9.

1 solution

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