A Problem on Infinite Sum and Recurrence Relations

Problem

Consider the infinite sum
$\mathbb{S} = \dfrac {a_0} {10^0} + \dfrac {a_1} {10^2} + \dfrac {a_2} {10^4} + .......$ where the sequence $\{a_n\}$ is defined by a_0=a_1=1 , and the recurrence relation $a_n=20a_{n-1} + 12 a_{n-2}$ for all positive integers $n \ge 2$ . If $\sqrt {\mathbb {S} }$ can be expressed in the form $\dfrac {a} {\sqrt{b}}$ where & are relatively prime positive integers. Determine the ordered pair $(a, \, b)$ .

Solution

As the recurrence relation states $a_n-20a_{n-1} - 12 a_{n-2} =0$ , we shall reform the infinite sum into the same pattern (have a deep look);
$\mathbb{S} - \dfrac {20 \mathbb{S} } {10^2} - \dfrac {12 \mathbb{S} } {10^4}$
$= ( \dfrac {a_0} {10^0} + \dfrac {a_1} {10^2} + \dfrac {a_2} {10^4} + \dfrac {a_3} {10^6} + .... )$ $- ( \dfrac {20a_0} {10^2} + \dfrac {20a_1} {10^4} + \dfrac {20a_2} {10^6} + \dfrac {20a_3} {10^8} + .... )$ $- ( \dfrac {12a_0} {10^4} + \dfrac {12a_1} {10^6} + \dfrac {12a_2} {10^8} + \dfrac {12a_3} {10^10} + .... )$

After Simplifying and arranging

$= \dfrac {a_0} {10^0} + \dfrac {a_1} {10^2} - \dfrac {20a_0} {10^2} + \dfrac {a_2 -20a_1-12a_0} {10^4} + \dfrac {a_3-20a_2-12a_1} {10^6} + \dfrac {a_4-20a_3-12a_2} {10^8} + . . . . . . \infty$
Now, as {the recurrence relation is}
$a_n-20a_{n-1}-12a_{n-2} =0$ for all $n \ge 2$ , all terms except first three are zero in R.H.S.
Hence we have,
$\mathbb{S} - \dfrac {20 \mathbb{S} } {10^2} - \dfrac {12 \mathbb{S} } {10^4} = \dfrac {a_0} {10^0} + \dfrac {a_1} {10^2} - \dfrac {20a_0} {10^2}$
and substituting a_0 =a_1=1 , we have
$\mathbb{S} - \dfrac {20 \mathbb{S} } {100} - \dfrac {12 \mathbb{S} } {10000}$
$= \dfrac {1} {1} + \dfrac {1} {100} - \dfrac {20} {100}$
or
$\dfrac {7988 \mathbb{S}} {10000} = \dfrac {81} {100}$
so,
$\mathbb{S} =2025/1997$
From the Problem,
$\sqrt {\mathbb{S}} = \sqrt{2025/1997} = 45/\sqrt{1997} =a/\sqrt{b}$
So, the desired ordered pair is (a, b) = (45, 1997) .