Răspunsuri

2014-11-09T19:06:35+02:00
Se aplica formula 1+a+ a^{2} + a^{2}+...+ a^{n}   \frac{ a^{n+1}-1 }{n-1}
S=  \frac{ 6^{2014}-1 }{5}
Dar probabil ca trebuie sa demonstrezi, asa ca:

S=1+6+6²+...+ 6^{2013}
6S=6+6²+6³+...+ 6^{2014}
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6S=S-1+ 6^{2014}
6S-S= 6^{2014}-1
5S= 6^{2014} -1
S= \frac{ 6^{2014}-1 }{5}
1 5 1