Remainders Are Fun 4

$\begin{aligned} a_n & = \frac {2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}} \\ \implies \frac 1{a_n} & = \frac 12 \left(\frac 1{a_{n-1}} + \frac 1{a_{n-2}} \right) & \small \color{#3D99F6} \text{Let }b_{n-1} = \frac 1{a_n} \\ 2b_{n-1} & = b_{n-2} + b_{n-3} & \small \color{#3D99F6} \text{Let }k=n-1 \\ 2b_k & = b_{k-1} + b_{k-2} \end{aligned}$

The characteristic equation of the linear recurrence relation of sequence $\{b_n\}$ is given below.

$\begin{aligned} 2r^2 - r - 1& = 0 \\ (2r+1)(r-1) & = 0 \\ \implies r & = - \frac 12, \ 1 \\ \implies b_k & = c_1\left(-\frac 12\right)^k + c_2 \\ b_0 & = \frac 1{a_1} \\ c_1 + c_2 & = \frac 1{20} \\ - \frac {c_1}2 + c_2 & = b_1 = \frac 1{a_2} = \frac 1{17} \\ \implies c_1 & = - \frac 1{170} \\ c_2 & = \frac {19}{340} \\ \implies b_k & = \frac {19}{340} - \frac 1{170}\left(-\frac 12\right)^k \end{aligned}$

Now we have:

$\begin{aligned} \sum_{k=1}^\infty \frac 1{2017^ka_k} & = \sum_{k=0}^\infty \frac {b_k}{2017^{k+1}} \\ & = \frac 1{2017} \left( \frac {19}{340}\sum_{k=0}^\infty \frac 1{2017^k} - \frac 1{170}\sum_{k=0}^\infty \left(\frac {-1}{4034}\right)^k\right) \\ & = \frac 1{2017} \left( \frac {19}{340}\left(\frac 1{1-\frac 1{2017}}\right) - \frac 1{170}\left(\frac 1{1+\frac 1{4034}}\right)\right) \\ & = \frac 1{2017} \left( \frac {19}{340}\left(\frac {2017}{2016}\right) - \frac 1{170}\left(\frac {4034}{4035}\right)\right) \\ & = \frac {19}{340\cdot 2016} - \frac 1{85\cdot 4035} \\ & = \frac {22867}{921916800} \end{aligned}$

$\begin{aligned} \implies X & = 20^2\cdot 2017^2\cdot 17 \cdot 135576 \cdot \sum_{k=1}^\infty \frac 1{2017^ka_k} \\ & = 20^2\cdot 2017^2\cdot 17 \cdot 135576 \cdot \frac {22867}{921916800} \\ & = 2017^2 \cdot 22867 \\ & = 17^2 \cdot 867 \text{ (mod 1000)} \\ & = 289(-133) \text{ (mod 1000)} \\ & = -437 \text{ (mod 1000)} \\ & = \boxed{563} \text{ (mod 1000)} \end{aligned}$

By Using the Iterative Formula Above, (This Solution Is So Hard and Long. )

$\begin{aligned} a_1 &= 20 \\ a_2 &= 17 \\ a_3 &= \dfrac{2(20)(17)}{20+17} = \dfrac{2^3 \cdot 5 \cdot 17}{37} \\ a_4 &= \dfrac{\dfrac{2^4 \cdot 5 \cdot 17^2}{37}}{17\left(\dfrac{37 + (2^3\cdot5)}{37}\right)} = \dfrac{2^4 \cdot 5 \cdot 17}{77} \\ a_5 &= \dfrac{\dfrac{2^8 \cdot 5^2 \cdot 17^2}{77\cdot37}}{2^3\cdot5\cdot17\left(\dfrac{77+(2\cdot37)}{37\cdot77}\right)} = \dfrac{2^5 \cdot 5 \cdot 17}{77+(2\cdot37)} \\ a_6 &= \dfrac{\dfrac{2^{10} \cdot 5^2 \cdot 17^2}{77\cdot(77+(2\cdot37))}}{2^4\cdot5\cdot17\left(\dfrac{77+(2\cdot37)+(2\cdot77)}{(77+(2\cdot37))\cdot77}\right)} = \dfrac{2^6 \cdot 5 \cdot 17}{77+(2\cdot37)+(2\cdot77)} \\ \vdots\end{aligned}$

$\displaystyle \sum_{k = 1}^{\infty} \dfrac{1}{2017^ka_k} = \dfrac{1}{2017\cdot20} + \dfrac{1}{2017^2\cdot17} + S$

where $S = \dfrac{37}{2017^3\cdot2^3\cdot85} + \dfrac{77}{2017^4\cdot2^4\cdot85} + \dfrac{77+(2\cdot37)}{2017^5\cdot2^5\cdot85} + \dfrac{77+(2\cdot37)+(2\cdot77)}{2017^6\cdot2^6\cdot85} + \cdots$

Multiplying $S$ by $4034$ and subtracting the result to $S$ yields:

$4033S = \dfrac{37}{2017^2\cdot2^2\cdot85} + \dfrac{40}{2017^3\cdot2^3\cdot85} + 2\left(\dfrac{S}{4034}\right) \implies \dfrac{8134560}{2017}S = \dfrac{1}{4034^2\cdot85} \cdot \dfrac{74649}{2017} \\ \implies S = \dfrac{24883}{4034^2\cdot85\cdot2711520} \\ \displaystyle \implies \sum_{k = 1}^{\infty} \dfrac{1}{2017^ka_k} = \dfrac{1}{2017} \left( \dfrac{1}{20} + \dfrac{1}{2017\cdot17} + \dfrac{24883}{8068\cdot85\cdot20\cdot135576}\right)$

$\begin{aligned} \implies 20^2\cdot 2017^2\cdot 17 \cdot 135576 \cdot \dfrac{1}{2017} \left( \dfrac{1}{20} + \dfrac{1}{2017\cdot17} + \dfrac{24883}{8068\cdot85\cdot20\cdot135576}\right) &= \begin{pmatrix} 20\cdot2017\cdot17\cdot135576 \\ + 20\cdot20\cdot135576 \\ + 24883 \end{pmatrix} \bmod 1000 \\ &\equiv \begin{pmatrix} 20\cdot17\cdot17\cdot576 \\ + 20\cdot20\cdot576 \\ + 883 \end{pmatrix} \\ &\equiv \begin{pmatrix} 280 \\ + 400 \\ + 883 \end{pmatrix} \equiv \ \boxed{563} \end{aligned}$