Question

Tìm n thuộc N

x²y+2x²y+3x²y+...+nx²y=210x²y

Answer 1

\(x^2y+2x^2y+3x^2y+...+nx^2y=210x^2y\)

\(\Leftrightarrow x^2y\left(1+2+3+....+n\right)=210x^2y\)

\(\Leftrightarrow1+2+3+....+n=210\)

\(\Leftrightarrow\frac{n\left(n+1\right)}{2}=210\)

\(\Leftrightarrow n\left(n+1\right)=420=20.21\)

\(\Rightarrow n=20\)

Answer 2

Ta có : x²y + 2x²​y + 3x²​y + ......+ nx²​y = 210x²​y

<=> x²​y(1 + 2 + 3 + ...... + n) = 210x²​y

=> 1 + 2 + 3 + ...... + n = 210

Áp dụng công thức \(\frac{n\left(n+1\right)}{2}=210\)

Mà 210 = 20.21

Nên n = 20