If x0,x1,x2,... is a sequence of numbers such thatx0=1000, and xn=−1000n(x0+x1+x2+...xn−1) for all n≥1. then the sum of digits of value of 122x0+12x1+x2+2x3+22x4+...+2997x999+2998x1000 is equal to_____.

Observe that when n≥2, { nxn=−1000(x0+x1+...+xn−1),(n−1)xn−1=−1000(x0+x1+...+xn−2).Thus, we have nxn−(n−1)xn−1=−1000xn−1⇒xn=−(1000−(n−1)n)xn−1.It is easy to check that the above formula holds even when n=1.Therefore, for 1≤n≤100, we have xn=−1000−(n−1)nxn−1 =(−1)21000−(n−1)n.1000−(n−2)n−1xn−2=... =(−1)n1000−(n−1)n.1000−(n−2)n−1...10001x0 =(−1)n(1000n)x0. Hence we have 122x0+12x1+x2+....+2998x1000 =14(x0+2x1+22x2+...+21000x1000)14(x0−(10001)2x0+(10002)22x0−(10003)23x0+...+(10001000)21000x0)=x04(1−(10001)2+(10002)22−(10002)23+...+(10001000)21000)=10004(1−2)1000=250.

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If $x_{0}, x_{1}, x_{2}, ...$ is a sequence of numbers such that $x_{0} = 1000,$ and $x_{n} = - \frac{1000}{n} (x_{0} + x_{1} + x_{2} + ... x_{n - 1})$
for all $n \geq 1.$ then the sum of digits of value of $\frac{1}{2^{2}} x_{0} + \frac{1}{2} x_{1} + x_{2} + 2 x_{3} + 2^{2} x_{4} + ... + 2^{997} x_{999} + 2^{998} x_{1000}$
is equal to_____.

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Detailed Solution

Observe that when $n \geq 2,$
   ${\begin{matrix} n x_{n} = - 1000 (x_{0} + x_{1} + ... + x_{n - 1}), \\ (n - 1) x_{n - 1} = - 1000 (x_{0} + x_{1} + ... + x_{n - 2}) . \end{matrix}$
Thus, we have
$n x_{n} - (n - 1) x_{n - 1} = - 1000 x_{n - 1} \Rightarrow x_{n} = - (\frac{1000 - (n - 1)}{n}) x_{n - 1} .$
It is easy to check that the above formula holds even when $n = 1.$ Therefore, for $1 \leq n \leq 100,$ we have
   $x_{n} = - \frac{1000 - (n - 1)}{n} x_{n - 1}$
   $= {(- 1)}^{2} \frac{1000 - (n - 1)}{n} . \frac{1000 - (n - 2)}{n - 1} x_{n - 2}$

$= ...$
   $= {(- 1)}^{n} \frac{1000 - (n - 1)}{n} . \frac{1000 - (n - 2)}{n - 1} ... \frac{1000}{1} x_{0}$
   $= {(- 1)}^{n} (\begin{matrix} 1000 \\ n \end{matrix}) x_{0} .$

Hence we have $\frac{1}{2^{2}} x_{0} + \frac{1}{2} x_{1} + x_{2} + .... + 2^{998} x_{1000}$

$\begin{array}{l} = \frac{1}{4} (x_{0} + 2 x_{1} + 2^{2} x_{2} + ... + 2^{1000} x_{1000}) \\ \frac{1}{4} (x_{0} - (\begin{matrix} 1000 \\ 1 \end{matrix}) 2 x_{0} + (\begin{matrix} 1000 \\ 2 \end{matrix}) 2^{2} x_{0} - (\begin{matrix} 1000 \\ 3 \end{matrix}) 2^{3} x_{0} + ... + (\begin{matrix} 1000 \\ 1000 \end{matrix}) 2^{1000} x_{0}) \\ = \frac{x_{0}}{4} (1 - (\begin{matrix} 1000 \\ 1 \end{matrix}) 2 + (\begin{matrix} 1000 \\ 2 \end{matrix}) 2^{2} - (\begin{matrix} 1000 \\ 2 \end{matrix}) 2^{3} + ... + (\begin{matrix} 1000 \\ 1000 \end{matrix}) 2^{1000}) \\ = \frac{1000}{4} {(1 - 2)}^{1000} \\ = 250. \end{array}$