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Moomo
a fost răspuns

Ultima cifra a n=1+2^0+2^1+2^2+2^3+2^4+...+2^2012 este: a. 8 b. 6 c. 4 d. 2

Răspuns :

Nicoletastefan2004
n-1=2^0+2^1+...+2^2012
2(n-+)=2^1+2^2+...+2^2012+2^2013
*le scadem*
n-1=2^2013-2^0
n-1=2^2013-1
n=2^2013
ulrima cifra=(2^4)^503+2^1=(...6)^503+...2=...6+...2=...8

raspunsul corect este a.8


Sper ca te am ajutat!:)
GeneralRex121

[tex]u(2^{1}) = 2 \\u(2^{2}) = 4 \\u(2^{3}) = 8 \\u(2^{4}) = 6 \\u(2^{2012}) = 6\\n - 2 = 2^{1} + 2^{2} + ... + 2^{2012}\\ u(n-2) = 0\\u(n) = 2[/tex]

503 grupe de 2 + 4 + 6 + 8 u(2+4+6+8) = 0 deci u(n-2) = 0;

u(n) = 2;