Question

[Lê Đức Anh]

Tìm y, biết: (y+1)+(y+2)+(y+3)+...+(y+99)=2550

Answer 1

[Nguyễn Tấn Phát]

\(\left(y+1\right)+\left(y+2\right)+\left(y+3\right)+...+\left(y+99\right)=2550\)

Có \(\left(99-1\right):1+1=99\)cặp

\(\left(y+y+y+...+y\right)+\left(1+2+3+...+99\right)=2550\)

\(99y+\left[\left(99+1\right)\times99:2\right]=2550\)

\(99y+4590=2550\)

\(99y=2550-4590\)

\(99y=-2040\)

\(y=\frac{-680}{33}\)

Đề hình như sai bạn ơi

Answer 2

[T.Ps]

#)Giải :

\(\left(y+1\right)+\left(y+2\right)+\left(y+3\right)+...+\left(y+99\right)=2550\)

\(\Rightarrow\left(y+y+y+...+y\right)+\left(1+2+3+...+99\right)=2550\)

\(\Rightarrow99y+\frac{\left(99+1\right).99}{2}=2550\)

\(\Rightarrow99y+4950=2550\)

\(\Leftrightarrow99y=-2400\)

\(\Leftrightarrow y=-24,2424242424.................\approx-24,24\)

P/s : Có j đó sai sai :v đề lỗi ak ?