Question

Tính tổng : S= 2016 + \(\frac{20\text{1}6}{\text{1}+2}\) + \(\frac{20\text{1}6}{\text{1}+2+3}\) +....+ \(\frac{20\text{1}6}{\text{1}+2+3+...+20\text{1}5}\)

Answer 1

\(S=\frac{2016}{2.3:2}+\frac{2016}{3.4:2}+...+\frac{2016}{2015.2016:2}\)

\(S=\frac{4032}{2.3}+\frac{4032}{3.4}+...+\frac{4032}{2015.2016}\)

\(S=4032\left[\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\right]\)

\(S=4032\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right]\)

\(S=4032\left[\frac{1}{2}-\frac{1}{2016}\right]=4032\cdot\frac{1007}{2016}\)

\(S=2014\)

Answer 2

S = \(2016+\frac{2016}{1+2}+\frac{2016}{1+2+3+}+...+\frac{2016}{1+2+3+...+2015}\)

S = \(2016+\left(\frac{2016}{1+2}+\frac{2016}{1+2+3}+...+\frac{2016}{1+2+3+...+2015}\right)\)

S = \(2016+2016.\left(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2015}\right)\)

đặt A = \(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2015}\)

A = \(\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2015\right).2015:2}\)

A = \(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2015.2016}\)

A = \(2.\left(\frac{1}{2}-\frac{1}{3}\right)+2.\left(\frac{1}{3}-\frac{1}{4}\right)+...+2.\left(\frac{1}{2015}-\frac{1}{2016}\right)\)

A = \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)

A = \(2.\left(\frac{1}{2}-\frac{1}{2016}\right)\)

A = \(2.\frac{1007}{2016}=\frac{1007}{1008}\)

Thay A vào ta được :

S = \(2016+2016.\frac{1007}{1008}\)

S = \(2016.\left(1+\frac{1007}{1008}\right)\)

S = \(2016.\frac{2015}{1008}\)

S = \(4030\)

Answer 3

Đào Trọng Luân    sai nha bạn

Answer 4

À Băng Di ơi mình thếu bước cộng 2016 vào, bạn cộng vào nha

S = 2016 + 2014 = 4030