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Calculeaza:
S=[tex] \sqrt{2} + \sqrt{ 2^{2} } + \sqrt{ 2^{3} } + ... + \sqrt{ 2^{100} } [/tex]
Repede va rog! :)


Răspuns :

S=√2+2+2√2+4+4√2+..............................2^49 +2^49√2+ .2^50=
S=2+2²+2³+..............+2^49+2^50 +√2(1+2+4+....2^48+2^49)=
S=2 ( 1+2+2²+...........+2^49) +√2(1+2+...+2^49);
dar
1+2+2²+...........+2^49 =2^50-1/ (2-1)=2^50-1
deci
S= (2+√2)(2^50-1)

altefel
S=√2 ( 1+√2+(√2)²+(√2)³+......(√2)^99)
 dar
( 1+√2+(√2)²+(√2)³+......(√2)^99)=[(√2)^100-1]/ (√2-1)= (aratioinalizam numitorul , prin amplificare cu comnjugata)
 √2(√2+1)(2^50-1)/(2-1)= (2+√2)(2^50-1) acelasi rezultat

ca sa il scriem cu parte rationala si irationala:

S=2 (2^50-1) +√2(2^50-1)

sau 2^51-2 +(2^50-1)√2