đại số lớp 7

B

baochauhn1999

Ta có:
$2^{2^{2n+1}}=2^{2.2^{2n}}=4^{2^{2n}}$
Xét:
$2^{2n}=4^n \equiv 1^n(mod 3) \equiv 1(mod 3)$
Suy ra: $4^{2^{2n}}=4^{3k+1}=4.64^{k}$
Cho nên:
$2^{2^{2n+1}}+3=4.64^{k}+3 \equiv 4.1^{k}+3(mod 7) \equiv 4+3(mod 7) \equiv 0(mod 7)$
Như vậy: $2^{2^{2n+1}}+3$ chia hết cho 7 hay $2^{2^{2n+1}}+3$ là hợp số.
 
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