Răspuns :
(....(((2+2)·2+2³)·2+2⁵)·2+...+2²⁰⁰¹)·2+2²⁰⁰³=
=(....(((2²·2+2³)·2+2⁵)·2+...+2²⁰⁰¹)·2+2²⁰⁰³=
=(....(((2³+2³)·2+2⁵)·2+...+2²⁰⁰¹)·2+2²⁰⁰³=
=(....((2³(2⁰+2⁰)·2+2⁵)·2+...+2²⁰⁰¹)·2+2²⁰⁰³=
=(....((2³(1+1)·2+2⁵)·2+...+2²⁰⁰¹)·2+2²⁰⁰³=
=(....((2³·2·2+2⁵)·2+...+2²⁰⁰¹)·2+2²⁰⁰³=
=(....((2⁵+2⁵)·2+...+2²⁰⁰¹)·2+2²⁰⁰³=
=(....((2⁶·2+...+2²⁰⁰¹)·2+2²⁰⁰³=
=(....((2⁷+...+2²⁰⁰¹)·2+2²⁰⁰³=
observam ca de fiecare data obtinem in paranteza centrala 2 la puterea n+1
deci calculand in continuare vom obtine
=(2²⁰⁰¹+2²⁰⁰¹)·2+2²⁰⁰³=
=2²⁰⁰²·2+2²⁰⁰³=
=2²⁰⁰³+2²⁰⁰³=
=2²⁰⁰⁴
S=1+2+2²+2³+....+2²⁰¹¹ inmultim cu 2
2S=2+2²+2³+2⁴....+2²⁰¹² scadem S
2S-S=2+2²+2³+2⁴....+2²⁰¹²-(1+2+2²+2³+....+2²⁰¹¹)
S=2+2²+2³+2⁴....+2²⁰¹²-1-2-2²-2³-....-2²⁰¹¹
S=2²⁰¹²-1
S=1+3+3²+3³+....+3²⁰⁰³ inmultim cu 3
3S=3+3²+3³+3⁴....+3²⁰⁰⁴ scadem S
3S-S=3+3²+3³+3⁴....+3²⁰⁰⁴-(1+3+3²+3³+....+3²⁰⁰³)
2S=3+3²+3³+3⁴....+3²⁰⁰⁴-1-3-3²-3³-....-3²⁰⁰³
2S=3²⁰⁰⁴-1
S=(3²⁰⁰⁴-1)/2
=(....(((2²·2+2³)·2+2⁵)·2+...+2²⁰⁰¹)·2+2²⁰⁰³=
=(....(((2³+2³)·2+2⁵)·2+...+2²⁰⁰¹)·2+2²⁰⁰³=
=(....((2³(2⁰+2⁰)·2+2⁵)·2+...+2²⁰⁰¹)·2+2²⁰⁰³=
=(....((2³(1+1)·2+2⁵)·2+...+2²⁰⁰¹)·2+2²⁰⁰³=
=(....((2³·2·2+2⁵)·2+...+2²⁰⁰¹)·2+2²⁰⁰³=
=(....((2⁵+2⁵)·2+...+2²⁰⁰¹)·2+2²⁰⁰³=
=(....((2⁶·2+...+2²⁰⁰¹)·2+2²⁰⁰³=
=(....((2⁷+...+2²⁰⁰¹)·2+2²⁰⁰³=
observam ca de fiecare data obtinem in paranteza centrala 2 la puterea n+1
deci calculand in continuare vom obtine
=(2²⁰⁰¹+2²⁰⁰¹)·2+2²⁰⁰³=
=2²⁰⁰²·2+2²⁰⁰³=
=2²⁰⁰³+2²⁰⁰³=
=2²⁰⁰⁴
S=1+2+2²+2³+....+2²⁰¹¹ inmultim cu 2
2S=2+2²+2³+2⁴....+2²⁰¹² scadem S
2S-S=2+2²+2³+2⁴....+2²⁰¹²-(1+2+2²+2³+....+2²⁰¹¹)
S=2+2²+2³+2⁴....+2²⁰¹²-1-2-2²-2³-....-2²⁰¹¹
S=2²⁰¹²-1
S=1+3+3²+3³+....+3²⁰⁰³ inmultim cu 3
3S=3+3²+3³+3⁴....+3²⁰⁰⁴ scadem S
3S-S=3+3²+3³+3⁴....+3²⁰⁰⁴-(1+3+3²+3³+....+3²⁰⁰³)
2S=3+3²+3³+3⁴....+3²⁰⁰⁴-1-3-3²-3³-....-3²⁰⁰³
2S=3²⁰⁰⁴-1
S=(3²⁰⁰⁴-1)/2