number properties

• February 26th 2009, 06:05 AM
Pinky9
number properties

Teachers doesn't ask a specific question but requires an investigation into this...anyone have an idea where to start? I have tried using different types of denominators as well as lower or higher numerators but no recurring pattern emerges.
• February 26th 2009, 06:45 AM
tah
$\frac{a+b(1+n+n^2+\cdots+n^{k-1})}{n^k}$
• February 26th 2009, 06:47 AM
Soroban
Hello, Pinky9!

Quote:

$a$ add $b$, divide by $n$,
. . . add $b$, divide by $n$,
. . . add $b$, divide by $n$ . . . recurring.

With no instructions, I will assume we want the ${\color{blue}k^{th}}$ term of the sequence.

Crank out the first few terms and look for a pattern . . .
. . Note that $n \neq 0$

. . $u_1 \:=\:\frac{a+b}{n}$

. . $u_2 \:=\:\frac{\frac{a+b}{n} + b}{n} \:=\:\frac{a+b+bn}{n^2}$

. . $u_3 \:=\:\frac{\frac{a+b+bn}{n} + b}{n} \;=\;\frac{a+b+bn + bn^2}{n^3}$

. . $u_4 \:=\:\frac{\frac{a+b+bn+bn^2}{n} + b}{n} \:=\:\frac{a+b+bn+bn^2+bn^3}{n^4}$

The $k^{th}$ term seems to be: . $u_k \;=\;\frac{a+b + bn + bn^2 + \hdots + bn^{k-1}}{n^k}$

. . $\text{and we have: }\:u_k \;=\;\frac{a + b\overbrace{(1 + n + n^2 + \hdots + n^{k-1})}^{\text{geometric series}}}{n^k}$

The geometric series has the sum: . $\frac{1-n^k}{1-n}$ . for $n \neq 1$
$\text{Hence: }\:u_k \;=\;\frac{1+b\,\frac{1-n^k}{1-n}}{n^k}$

$\text{Therefore: }\:\boxed{u_k \;=\;\frac{(1-n)a + (1-n^k)b}{n^k(1-n)}}$

• February 27th 2009, 12:13 AM
Pinky9
Thank you so much - I should have thought of that!