Simplifying Expression - Algebra

**Simplicity** · November 22nd 2009, 08:12 AM

In one of my textbook example, after the summation result was given, they simplified the expression but I can't see how this simplification was done.

$0.5^k\left[\frac{1-(0.8/0.5)^{k+1}}{1-(0.8/0.5)}\right]=\frac{10}{3}[0.8^{k+1}-0.5^{k+1}]$

I don't see how the LHS simplified to the RHS? All help will be appreciated. Thanks in advance.

**ukorov** · November 22nd 2009, 09:46 AM

not sure yet but looks like the denominator 1 - (0.8/0.5) may be taken care of using forumla (a - b)(a + b) = a^2 - b^2 ....

**Soroban** · November 22nd 2009, 09:51 AM

Hello, Simplicity!

$0.5^k \cdot \frac{1-\left(\dfrac{0.8}{0.5}\right)^{k+1}} {1-\left(<br /> \dfrac{0.8}{0.5}\right)} \;\;=\;\; (0.5)^k \cdot \frac{\dfrac{0.5^{k+1} - 0.8^{k+1}} {0.5^{k+1}}} {\dfrac{0.5-0.8}{0.5}}$

. . . $=\;\; (0.5)^k\cdot \frac{0.5^{k+1} - 0.8^{k+1}}{0,5^{k+1}}\cdot\frac{0.5}{0.5-0.8} \;\;=\;\;\frac{0.5^{k+1} - 0.8^{k+1}}{-0.3}$

. . . $=\;\; \frac{0.5^{k+1} - 0.8^{k+1}}{-\frac{3}{10}} \;\;=\;\;-\frac{10}{3}\left(0.5^{k+1} - 0.8^{k+1}\right) \;=\;\frac{10}{3}\left(0.8^{k+1}-0.5^{k+1}\right)$

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