f : R → R
f(x) = 2x-1
a= f(1) + f(2) + f(3) + ... + f(214)
a ∈ N, a ∈ p.p
f(1) = 2·1-1=1
f(2) = 2·2-1=3
f(3) = 2·3-1=5
f(4) = 2·4-1=7
...
f(214) = 2·214-1 = 427
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Observam ca toate numerele vor fi impare, deci aplicam urmatoarea regula:
a= 1+3+5+...+427 = S(1;427) - S(2-426)
[numere pare]
a= 1+2+3+4+5+...+427-2-4-6-...-426
a= [tex] \frac{427(427+1)}{2} [/tex] - 2(1+2+3+...+213)
a= [tex] \frac{427*428}{2} [/tex] - 2·[tex] \frac{213(213+1)}{2} [/tex]
simplific.....
a= 427·214 - 213·214
a= 91378 - 45582
a = 45796
a²= 45792
a = √45792
a = 214 → a = p.p
a ∈ N
