Question

cho n là STN

CM: 5²ⁿ⁺¹ + 2²ⁿ⁺⁴ + 2ⁿ⁺¹\(⋮\) 23

Answer 1

Có hằng đẳng thức: $a^n - b^n = (a-b)[a^{n-1}.b + a(n-2).b$² $+..+ b^(n-1)] = (a-b).p$

* $5^{2n} - 2^n = 25^n - 2^n = (25-2)p = 23p => 5.5^{2n} - 5.2^n = 5.23.p$
$=> 5^{2n+1} - 5.2^n = 5.23p$ chia hết cho 23

* $2^{n+4} + 2^{n+1} = 2^n.2^4 + 2^n.2 = 2^n(2^4 + 2) = 18.2^n = 23.2^n - 5.2^n $

Vậy: $5^{2n+1} + 2^{n+4} + 2^{n+1} = 5^{2n+1} - 5.2^n + 23.2^n$ chia hết cho 23