Câu hỏi của My Xu

Question

Tìm $n\in N$biết: $x^3y^4+2x^3y^4+3x^3y^4+...+nx^3y^4=820x^3y^4$

Đinh Đức Hùng · Accepted Answer

$x^3y^4+2x^3y^4+3x^3y^4+....+nx^3y^4=820x^3y^4$

$\Leftrightarrow x^3y^4\left(1+2+3+....+n\right)=820x^3y^4$

$\Leftrightarrow1+2+3+....+n=820$

$\Leftrightarrow\frac{n\left(n+1\right)}{2}=820$

$\Leftrightarrow n\left(n+1\right)=1640=40.41$

$\Rightarrow n=40$

Câu trả lời:

$x^3y^4+2x^3y^4+3x^3y^4+....+nx^3y^4=820x^3y^4$

$\Leftrightarrow x^3y^4\left(1+2+3+....+n\right)=820x^3y^4$

$\Leftrightarrow1+2+3+....+n=820$

$\Leftrightarrow\frac{n\left(n+1\right)}{2}=820$

$\Leftrightarrow n\left(n+1\right)=1640=40.41$

$\Rightarrow n=40$

Mạnh Lê · Answer

$x^3y^4+2x^3y^4+3x^3y^4+...+nx^3y^4=820x^3y$

$\Leftrightarrow x^3y^4\left(1+2+3+...+n\right)=820x^3y^4$

$\Leftrightarrow1+2+3+...+n=820$

$\Leftrightarrow\frac{n\left(n+1\right)}{2}=820$

$\Leftrightarrow n\left(n+1\right)=1640=40,61$

$n=40$

Câu trả lời:

$x^3y^4+2x^3y^4+3x^3y^4+...+nx^3y^4=820x^3y$

$\Leftrightarrow x^3y^4\left(1+2+3+...+n\right)=820x^3y^4$

$\Leftrightarrow1+2+3+...+n=820$

$\Leftrightarrow\frac{n\left(n+1\right)}{2}=820$

$\Leftrightarrow n\left(n+1\right)=1640=40,61$

$n=40$