Question

Giúp với mọi người ơi :v

Answer 1

\(A=1^5+2^5+3^5+......+100^5\)

Ta luôn có \(a^n+b^n=\left(a+b\right)\left(.....\right)\)   \(\text{∀}\) \(n\) ∈ \(N\) , \(n:\) \(lẻ\)

\(A=\left(1^5+99^5\right)+\left(2^5+98^5\right)+.....+\left(49^5+51^5\right)+100^5+50^5\)

\(A=\left(99+1\right)\left(...\right)+\left(2+98\right)\left(....\right)+....+\left(49+51\right)\left(....\right)+100^5+50^5\)

\(A=100\left(...\right)+\left(40+10\right)^5\)

\(Đặt\) \(P=\left(40+10\right)^5=40^5+10^5+5.40^4.10+10.40^3.10^2+10.40^2.10^3+5.40.10^4+10^5\)

Dễ thấy \(P\) ⋮ 4 \(=>A\) ⋮ 4 \(=>\) đpcm