Este o progresie geometrică cu n + 1 termeni.
Formula sumei este:
[tex]\boldsymbol{S_{n} = \dfrac{b_{1} \cdot \big(q^{n} - 1\big)}{q - 1}}[/tex]
[tex]b_{1} = 1, \ q = \dfrac{1}{2}[/tex]
[tex]x = S_{n+1} = \dfrac{1 \cdot \bigg(1 - \dfrac{1}{2^{n + 1}}\bigg)}{1 - \dfrac{1}{2}} = \dfrac{1 - \dfrac{1}{2^{n+1}}}{\dfrac{1}{2}} =\\[/tex]
[tex]= 2 \bigg(1 - \dfrac{1}{2^{n+1}}\bigg) = \bf2 - \dfrac{1}{2^{n}}[/tex]
