Question

Tìm n, biết:

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

Answer 1

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

  \(x^2y\left(1+2+3+4+...+n\right)=210x^2y\)

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

\(1+2+3+...+n=210\)

\(\frac{\left(n-1\right):1+1}{2}.\left(n+1\right)=210\)

\(n\left(n+1\right):2=210\)

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

vậy n=20

Answer 2

1+2+3+...+n = 210

n(n+1):2=210

n(n+1)=420 =20.21

n =20

Answer 3

n=20 nha rồi mình kik hộ cho