Q37 / 90 · KVPY 2017 (19 Nov SB/SX) · Mathematics
Let $\mathrm{S}$ be the infinite sum given by $\mathrm{S}=\sum_{\mathrm{n}=0}^{\infty} \frac{\mathrm{a}_{\mathrm{n}}}{10^{2 \mathrm{n}}}$
where $\left\{a_{n}\right\}_{n \geq 0}$ is a sequence defined by $a_{0}=a_{1}=1$ and $a_{j}=20 a_{j-1}-108 a_{j-2}$ for $j \geq 2$.
If $S$ is expressed in the form $\frac{a}{b}$, where $a, b$ are coprime positive integers, then a equals
- A2017
- B2020
- C2023
- Correct answerD2025
Solution
$\begin{array}{l}
a_{n}=20 a_{n-1}-108 a_{n-2} \\
\frac{a_{n}}{10^{2 n}}=\frac{20 a_{n-1}}{10^{2 n}}-\frac{108 a_{n-2}}{10^{2 n}} \\
\frac{a_{n}}{10^{2 n}}=\frac{20}{100} \frac{a_{n-1}}{10^{2(n-1)}}-\frac{108}{10000} \frac{a_{n-2}}{10^{2(n-2)}}
\end{array}$
apply summation
$\begin{array}{l}
\sum_{\mathrm{n}=2}^{\infty} \frac{\mathrm{a}_{\mathrm{n}}}{10^{2 \mathrm{n}}}=\frac{1}{5} \sum_{\mathrm{n}=2}^{\infty} \frac{\mathrm{a}_{\mathrm{n}-1}}{10^{2(\mathrm{n}-1)}}-\frac{27}{2500} \sum_{\mathrm{n}=2}^{\infty} \frac{\mathrm{a}_{\mathrm{n}-2}}{10^{2(\mathrm{n}-2)}} \\
\mathrm{S}-1-\frac{1}{100}=\frac{1}{5}(\mathrm{~S}-1)-\frac{27}{2500} \mathrm{~S} \\
\mathrm{~S}-1-\frac{1}{100}=\frac{1}{5} \mathrm{~S}-\frac{1}{5}-\frac{27}{2500} \mathrm{~S} \\
\mathrm{~S}\left(1-\frac{1}{5}+\frac{27}{2500}\right)=-\frac{1}{5}+1+\frac{1}{100}
\end{array}$
$\mathrm{~S}\left(\frac{2500-500+27}{2500}\right)=\frac{81}{100}$
$\mathrm{~S}=\frac{81 \times 25}{2027}$
$\mathrm{~S}=\frac{2025}{2027}$
Hence, the correct answer is option: 'D'.