2/(1/1+2+1/1+2+3+...+1/1+2+3+...+2015) = ?
Tính S = 1/2(1+2) + 1/3(1+2+3)+...+ 1/2015(1+2+...+2014+2015) + 1/2016(1+2+...+2015+2016)
Tinh:
S=2015 + 2015/1+2 +2015/1+2+3 + 2015/1+2+3+4 +... + 2015/1+2+3+...+2016
Tinh:
S=2015 + 2015/1+2 +2015/1+2+3 + 2015/1+2+3+4 +... + 2015/1+2+3+...+2016
tìm s
Tính S = 1/2(1+2) + 1/3(1+2+3)+...+ 1/2015(1+2+...+2014+2015) + 1/2016(1+2+...+2015+2016
Tính C= 1+1/2(1+2)+1/3(1+2+3)+........+1/2015(1+2+3+4+...+2015)
C = 1+1/2(1+2)+1/3(1+2+3)+........+1/2015(1+2+3+4+...+2015)
C = 1 + \(\frac{1}{2}\cdot\frac{2.3}{2}\)+ \(\frac{1}{3}\cdot\frac{3.4}{2}\)+ ... + \(\frac{1}{2015}\cdot\frac{2015.2016}{2}\)
C = \(\frac{2}{2}\) + \(\frac{3}{2}+\frac{4}{2}+...+\frac{2016}{2}\)
C = \(\frac{2+3+4+...+2016}{2}\)
Đặt D = 2 + 3 + 4 + ... + 2016
Số số hạng của D là : (2016 - 2) : 1 + 1 = 2015
Tổng D là : (2 + 2016) . 2015 : 2 = 2033135
Thay D vào biểu thức C ta được : \(\frac{2033135}{2}\)
Vậy C = ... .
Tính ;1+1/2×(1+2)+1/3×(1+2+3)+….+1/2015×(1+2+…+2015)
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{2015}\left(1+2+3+...+2015\right)\)
\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+...+\frac{1}{2015}.2015.2016:2\)
\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{2016}{2}=\frac{2+3+4+...+2016}{2}=\frac{2033135}{2}\)
1+1/2.(1+2)+1/3.(1+2+3)+............+1/2015.(1+...2015) = ?
2015 + (2015/ 1+2) + (2015/ 1+2+3) +......+ (2015/ 1+2+3+...+2014) =?
Tính
1+1/2×(1+2)+1/3×(1+2+3)+….+1/2015×(1+2+…+2015)
A=1+1/2(1+2)+1/3(1+2+3)+...+1/2015(1+2+...+2015)
tính A