Răspuns:
2²⁰⁰⁸ - 2¹⁰
Explicație pas cu pas:
[tex]A = {2}^{10} + {2}^{11} + {2}^{12} + ... + {2}^{2007} = \\ = {2}^{10} \cdot (1 + {2}^{1} + {2}^{2} + ... + {2}^{1997}) = {2}^{10} \cdot S[/tex]
[tex]S = 1 + {2}^{1} + {2}^{2} + ... + {2}^{1997} \\ [/tex]
[tex]2S = {2}^{1} + {2}^{2} + ... + {2}^{1997} + {2}^{1998} \\ [/tex]
[tex]2S + 1 = 1 + {2}^{1} + {2}^{2} + ... + {2}^{1997} + {2}^{1998} \\ [/tex]
[tex]2S + 1 = S + {2}^{1998} \iff S = {2}^{1998} - 1\\ [/tex]
[tex]A = {2}^{10} \cdot ({2}^{1998} - 1) \implies \red {\bf A = {2}^{2008} - {2}^{10}} \\ [/tex]
