1. ## geometric progressions

show that the sum of n terms of the series

$\log a + \log (2a) + \log (4a) + \log (8a) + ....$

is $\log [a^{n} \, 2^{n(n-1)/2}]$

2. Originally Posted by Rose Wanjohi
show that the sum of n terms of the series

$\log a + \log (2a) + \log (4a) + \log (8a) + ....$

is $\log [a^{n} \, 2^{n(n-1)/2}]$
Start by noting that the series can be written as

$\log a + (\log a + \log 2 ) + (\log a + 2 \log 2 ) + (\log a + 3 \log 2 ) + ....$

$= n \log a + (1 + 2 + 3 + .... + n) \log 2$.

The log rule $\log A^B = B \log A$ wil also be useful.

3. Hello, Rose!

Show that: . $\log a + \log (2a) + \log (4a) + \log (8a) + \hdots \log(2^{n-1}a) \;=\;\log \left[a^{n} \, 2^{\frac{n(n-1)}{2}}\right]$

$\log a + \log(2a) + \log(4a) + \log(8a) + \hdots + \log(2^{n-1}a)$

. . $= \;\log\bigg[a\cdot2a\cdot4a\cdot8a\cdots 2^{n-1}a\bigg]$

. . $= \;\log\bigg[ ( 2\cdot4\cdot8 \cdots 2^{n-1}) (a\cdot a\cdot a\cdot a \cdots a)\bigg]$

. . $= \;\log\bigg[\left(2\cdot2^2\cdot2^3 \cdots 2^{n-1}\right)(a^n)\bigg]$

. . $= \;\log\bigg[ \left(2^{1+2+3+\hdots + (n-1)}\right)\left(a^n \right)\bigg]$

. . $= \;\log\bigg[2^{\frac{n(n-1)}{2}}a^n\bigg]$