nhân 3 lên rồi tính

Đặt \(A=1.2+2.3+...+59.60\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+...+59.60.\left(61-58\right)\)

\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+...+59.60.61-58.59.60\)

\(\Rightarrow3A=59.60.61\)

\(\Rightarrow A=\frac{59.60.61}{3}\)

Gọi tổng trên là A, ta có:

A=1x2+2x3+...+59x60

3.A=1x2x3+2x3x3+...+59x60x3

3.A=1x2x3+2x3x(4-1)+...+59x60x(61-58)

3.A=(1x2x3+2x3x4+...+59x60x61)-(1x2x3+2x3x4+...+57x58x59)

3.A=59x60x61

A=59x20x61

A=71980

chúc bạn học tốt nha

Trả lời

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}\)

\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}< 1\)

Vậy \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}< 1\left(đpcm\right)\)

1-1/50=49/50<1

bạn viết sai đề rồi

1/2x3+1/3x4+1/4x5+.......+1/57x58+1/58x59+1/59x60 

1/2-1/3+1/3-1/4 +1/4 - 1/5 +... + 1/59-1/60

=1/2-1/60

=29/60

\(=\frac{29}{60}\) tích nha !

c, A= 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/100-1/101

    A= 1-1/101

    A= 100/101

Vậy A= 100/101

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2019}-\frac{1}{2020}\)

\(=1-\frac{1}{2020}>1\)

Thank you bạn dcv new ^ ^

nhầm dấu rồi bé hơn 1 chứ:v bạn sửa lại hộ mình nhà

\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

\(=1-\dfrac{1}{25}\)

\(=\dfrac{24}{25}\)

Nhanh giúp mình với cả nhà ơi