Question

Tính giá trị biểu thức:

P= 1+\(\dfrac{1}{2}\)(1+2)\(\dfrac{1}{3}\)(1+2+3)+\(\dfrac{1}{4}\)(1+2+3+4)+...+\(\dfrac{1}{2016}\)(1+2+3+...+2016)

Answer 1

\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{2016}\left(1+2+...+2016\right)\)\(=1+\dfrac{2.3}{2.2}+\dfrac{3.4}{3.2}+...+\dfrac{2016.2017}{2016.2}\)

\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{2017}{2}\)

\(=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{2017}{2}\)

\(=\dfrac{1}{2}\left(2+3+...+2017\right)\)

Đặt \(A=2+3+...+2017\)

\(=2017+2016+...+2\)

\(\Rightarrow2A=\left(2+2017\right)+\left(3+2016\right)+...+\left(2017+2\right)\) ( 2016 cặp số )

\(\Rightarrow2A=2019+2019+...+2019\) ( 2016 số )

\(\Rightarrow2A=4070304\)

\(\Rightarrow A=2035152\)

\(\Rightarrow P=1017576\)

Vậy...

Answer 2

P= 1+1/2.3+1/3.6+...+1/2016.2033136

P= 1+3/2+2+...+2017/2

P= 2/2+3/2+4/2+...+2017/2

P=\(\dfrac{2+3+4+...+2017}{2}\)

P= \(\dfrac{2035152}{2}\)

P= 1017576

Answer 3

Anh Nguyễn Huy Tú làm đúng như hơi gọn;mình sẽ làm lại đầy đủ hơn nhé.

\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+....+\dfrac{1}{2016}\left(1+....+2016\right)\\ =1+\dfrac{1}{2}.\dfrac{\left(1+2\right).2}{2}+\dfrac{1}{3}.\dfrac{\left(1+3\right).3}{2}+...+\dfrac{1}{2016}.\dfrac{\left(2016+1\right).2016}{2}\\ =\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+....+\dfrac{2017}{2}=\dfrac{\left(2+3+....+2017\right)}{2}=\dfrac{\left(2017+2\right).2016}{2}=2009.1008=2025072\)