Câu hỏi của Thanh Quân

Question

Giúp với mọi người ơi :v

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Thanh Quân · Accepted Answer

$A=1^5+2^5+3^5+......+100^5$

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