help please!

The answer works out to 2002*3/2 = 3003.

Thanks!

Determine the value of \(2002 + \frac{1}{2} \left( 2001 + \frac{1}{2} \left( 2000 + \dots + \frac{1}{2} \left( 3 + \frac{1}{2} \cdot 2 \right) \right) \dotsb \right)\).

\(\begin{array}{|rcll|} \hline s&=& 2002 + \frac{1}{2} \left( 2001 + \frac{1}{2} \left( 2000 + \dots + \frac{1}{2} \left( 3 + \frac{1}{2} \cdot 2 \right) \right) \dotsb \right) \\ s &=& 2002 + 2001\left(\frac{1}{2}\right) + 2000 \left(\frac{1}{2}\right)^2+ \dots +3\left(\frac{1}{2}\right)^{1999} + 2 \left(\frac{1}{2}\right)^{2000} \\ && \boxed{ \text{We substitute }x =\frac{1}{2} } \\ s &=& 2002 + 2001x + 2000x^2+ \dots +3x^{1999} + 2x^{2000} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s &=& 2002 + & 2001x + 2000x^2+ \dots +3x^{1999} + 2x^{2000} \\ xs &=& & 2002x + 2001x^2 + \dots +4x^{1999} + 3x^{2000} + 2x^{2001} \\ \hline xs-s &=& &x + x^2 + \dots +x^{1999}+ x^{2000}+ 2x^{2001}-2002 \\ s(x-1) &=& & x\underbrace{(1+ x^2 + \dots +x^{1999})}_{\mathbf{= \frac{1-x^{2000}}{1-x}} }+ 2x^{2001}-2002 \\ s(x-1) &=& & \frac{ x(1- x^{2000})} { 1-x}+ 2x^{2001}-2002 \quad | \quad x =\frac{1}{2} \\\\ s(\frac{1}{2}-1) &=& & \frac{ \frac{1}{2} \left(1-\left(\frac{1}{2}\right)^{2000} \right)} {1-\frac{1}{2}}+ 2\left(\frac{1}{2}\right)^{2001}-2002 \\ -\frac{1}{2}s &=& & 1-\left(\frac{1}{2}\right)^{2000} + 2\left(\frac{1}{2}\right)^{2001}-2002 \\ -\frac{1}{2}s &=& & 1-\left(\frac{1}{2}\right)^{2000} + 2\left(\frac{1}{2}\right)^{2000}\frac{1}{2}-2002 \\ -\frac{1}{2}s &=& & 1-\left(\frac{1}{2}\right)^{2000} + \left(\frac{1}{2}\right)^{2000} -2002 \\ -\frac{1}{2}s &=& & 1 -2002 \\ \frac{1}{2}s &=& & 2002-1 \\ \frac{1}{2}s &=& & 2001 \\ \mathbf{s} &=& & \mathbf{4002} \\ \hline \end{array}\)

\(2002 + \frac{1}{2} \left( 2001 + \frac{1}{2} \left( 2000 + \dots + \frac{1}{2} \left( 3 + \frac{1}{2} \cdot 2 \right) \right) \dotsb \right) = \mathbf{4002}\)

a=2;b=2001;d=0;f=1;cycle:d=((d+a)/2);a++;if(a<=b, goto cycle,0);printd+2002

OUTPUT = 4,002