1+1/2×(1+2)+1/3×(1+2+3)+1/4×(1+2+3+4)+....+1/2017×(1+2+3+....+2017)
H24
Những câu hỏi liên quan
Tính S = 1 + 1/2.(1 + 2) + 1/3.(1 + 2 +3) + 1/4.(1 + 2 +3 + 4) +....+ 1/2017.(1 + 2 + 3 +...+ 2017)
Tính tổng \(S=\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+\frac{3}{3^4+3^2+1}+...+\frac{2017}{2017^4+2017^2+1}\)
Cho biểu thức : B = 2017+2017/1+2+2017/1+2+3+2017/1+2+3+4+....+2017/1+2+3+...+2012
Tính tổng \(S=\dfrac{1}{1^4+1^2+1}+\dfrac{2}{2^4+2^2+1}+\dfrac{3}{3^4+3^2+1}+...+\dfrac{2017}{2017^4+2017^2+1}\)
TÍnh
S=1+1/2(1+2)+1/3(1+2+3)+1/4+(1+2+3+4)+.....+1/2017(1+2+3+...+2017)
A= ( 1/2017+ 2/2016+ 3/2015+...+ 2015/3+ 2016/2+ 2017) : ( 1/2+1/3+1/4+...+1/2017+1/2018)
1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 +.....+1/2017×2017
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2017.2018}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{1}-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
Đúng 0
Bình luận (0)
= 1/1- 1/2+...+1/2017-1/2018
=1-1/2018
=2017/2018
k mk nha các bạn
Đúng 0
Bình luận (0)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=\frac{1}{1}-\frac{1}{2017}=\frac{2016}{2017}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Chứng minh rằng F= 1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+4+5+6+...+2017)<2016/2017
tính m=2016+2016/2+2015/3+2014/4+...+1/2017/1/2+1/3+1/4+...+1/2017