Tính 1+ 1/2+ 1/1+2+3+ 1/1+2+3+4+...+1/1+2+3+...+2016
Những câu hỏi liên quan
tính gt của biểu thức:
1+1/2.(1+2)+1/3.(1+2+3)+1/4.(1+2+3+4)+...+1/2016.(1+2+3+...+2016)
TÍNH GIÁ TRỊ BIỂU THỨC A=2016+(2016/1+2)+(2016/1+2+3)+....+(2016/1+2+3+4+...+2016)
tính a: [1-1/2*2]*[1-1/3*3]*[1-1/4*4]*[1-1/5*5]*......*[1-1/2015*2015]*[1-1/2016*2016]
Câu 1
a) Chứng tỏ rằng 1/3 - 1/3^2 + 1/3^3 - 1/3^4 + 1/3^5 - 1/3^6 < 1/4
b) Cho A= 2015^2016 + 2016^2015 x 2015 và B= 1 + 2^2 + 3^2 + ......+2016^2. Tính AB có chia hết cho 5 không? Vì sao?
tính \(C=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{2016}\left(1+2+3+...+2016\right)\)
\(C=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+..+\frac{1}{2016}.\left(1+2+3+...+2016\right)\)
\(C=1+\frac{1}{2}.\left(1+2\right).2:2+\frac{1}{3}.\left(1+3\right).3:2+\frac{1}{4}.\left(1+4\right).4:2+...+\frac{1}{2016}.\left(1+2016\right).2016:2\)
\(C=1+3:2+4:2+5:2+...+2017:2\)
\(C=2.\frac{1}{2}+3.\frac{1}{2}+4.\frac{1}{2}+5.\frac{1}{2}+...+2017.\frac{1}{2}\)
\(C=\frac{1}{2}.\left(2+3+4+5+...+2017\right)\)
\(C=\frac{1}{2}.\left(2+2017\right).2016:2\)
\(C=\frac{1}{2}.2019.2016.\frac{1}{2}\)
\(C=2019.504=1017576\)
Đúng 0
Bình luận (2)
tính m=2016+2016/2+2015/3+2014/4+...+1/2017/1/2+1/3+1/4+...+1/2017
Cho
A = 1/2 + 1/3 + 1/4 + ... + 1/2017
B = 1/2016 + 2/2015 +3/2014+ ...+ 2015/2 + 2016/1
Tính B : A
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
Đúng 1
Bình luận (0)
Tính: (2/3 + 3/4 + 4/5 + ... + 2016/2017) x (1/2 + 2/3 + 3/4 + ... + 2015/2016) – (1/2 + 2/3 + 3/4 + ... + 2016/2017) x (2/3 + 3/4 + 4/5 + ... + 2015/2016).
Cách 1:
Xét số bị trừ, ta có:
(2/3 + 3/4 + 4/5 + ... + 2016/2017) x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
= (2/3 + 3/4 + 4/5 + ... + 2015/2016 + 2016/2017) x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
= (2/3 + 3/4 + 4/5 + ... + 2015/2016) x (1/2 + 2/3 + 3/4 + ... + 2015/2016) + 2016/2017 x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
Xét số trừ, ta có:
(1/2 + 2/3 + 3/4 + ... + 2016/2017) x (2/3 + 3/4 + 4/5 + ... + 2015/2016)
= (1/2 + 2/3 + 3/4 + ... + 2015/2016 + 2016/2017) x (2/3 + 3/4 + 4/5 + ... + 2015/2016)
= (1/2 + 2/3 + 3/4 + ... + 2015/2016) x (2/3 + 3/4 + 4/5 + ... + 2015/2016) + 2016/2017 x (2/3 +3/4 + 4/5 + ... + 2015/2016) =
Ta thấy số bị trừ và số trừ có số hạng giống nhau là:
(2/3 +3/4 + 4/5 + ... + 2015/2016) x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
Nên phép trừ trên có thể viết lại:
2016/2017 x (1/2 + 2/3 + 3/4 + ... + 2015/2016) - 2016/2017 x (2/3 + 3/4 + 4/5 + ... + 2015/2016)
= 2016/2017 x [(1/2 + 2/3 + 3/4 + ... + 2015/2016) - (2/3 +3/4 + 4/5 + ... + 2015/2016)]
= 2016/2017 x 1/2
= 1008/2017
Cách 2:
Đúng 0
Bình luận (0)
Tính: (2/3 + 3/4 + 4/5 + ... + 2016/2017) x (1/2 + 2/3 + 3/4 + ... + 2015/2016) – (1/2 + 2/3 + 3/4 + ... + 2016/2017) x (2/3 + 3/4 + 4/5 + ... + 2015/2016).
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Tính :
\(I=\frac{2:2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.....+\frac{1}{1+2+3+......+2016}}\)