a)
O sa gasim un mod de a calcula urmatoarea suma; forma generala a lui a si b:
[tex]S=1+ \frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} +...+\frac{1}{2^{x-1}}+ \frac{1}{2^x}\\\\
\text{Inmultim cu } \frac{1}{2} \text{ in ambii membri:}\\\\
\frac{1}{2}S=1\cdot\frac{1}{2}+ \frac{1}{2^1} \cdot \frac{1}{2} + \frac{1}{2^2} \cdot \frac{1}{2} +...+ \frac{1}{2^{x-1}} \cdot \frac{1}{2} + \frac{1}{2^x} \cdot \frac{1}{2}\\\\
\frac{1}{2}S= \underbrace{\frac{1}{2^1}+ \frac{1}{2^2}+ \frac{1}{2^3} +...+ \frac{1}{2^x}}_{S-1} + \frac{1}{2^{x+1}} \\\\
\frac{1}{2}S=S-1+ \frac{1}{2^{x+1}}\\\\
[/tex]
[tex]S=2S-2+ \frac{2}{2^{x+1}}\\
S=2- \frac{2}{2^x+1}=\boxed{2- \frac{1}{2^x} }[/tex]
Aceea este formula, asadar, in cazul lui a, x este 4 ==> a = 2 - 1 / (2^4)
b)
[tex]b-a=(1+ \frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4}+\frac{1}{2^5})-(1+ \frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4})\\
b-a=1-1+\frac{1}{2^1} -\frac{1}{2^1} + \frac{1}{2^2}- \frac{1}{2^2}+\frac{1}{2^3} -\frac{1}{2^3} + \frac{1}{2^4}- \frac{1}{2^4}+\frac{1}{2^5}\\
b-a=\frac{1}{2^5}[/tex]
Din formula de la a) stim ca:
[tex]b=2- \frac{1}{2^5} \\
b+\frac{2}{2^5}=2-\frac{1}{2^5}+ \frac{2}{2^5}=2+\frac{1}{2^5} [/tex]
Din formula, se observa ca b < 2
