Question

[___Vương Tuấn Khải___]

Tính giá trị của biểu thức sau:

A = \(\dfrac{1}{2}\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{2015.2017}\right)\)

Answer 1

[Akai Haruma]

Lời giải:

Xét tổng quát:

\(1+\frac{1}{k(k+2)}=\frac{k(k+2)+1}{k(k+2)}=\frac{(k+1)^2}{k(k+2)}\)

Thay $k=1,2,....,2015$ ta có:

\(1+\frac{1}{1.3}=\frac{2^2}{1.3}\)

\(1+\frac{1}{2.4}=\frac{3^2}{2.4}\)

\(1+\frac{1}{3.5}=\frac{4^2}{3.5}\)

\(1+\frac{1}{4.6}=\frac{5^2}{4.6}\)

.............

\(1+\frac{1}{2015.2017}=\frac{2016^2}{2015.2017}\)

Nhân theo vế:

\(\Rightarrow A=\frac{1}{2}\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{2015.2017}\right)\)

\(=\frac{1}{2}.\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}....\frac{2016^2}{2015.2017}\)

\(=\frac{(1.2.3...2016)^2}{(1.2.3...2015)(2.3.4...2017)}=\frac{(1.2.3...2016)(2.3....2016)}{(1.2.3...2015)(2.3.4...2017)}=2016.\frac{1}{2017}=\frac{2016}{2017}\)