=[1*(1+2)/(1+2)-1/(1+2)]*...*[1*(1+2+3+...+2014)/(1+2+3+...+2014)-1/(1+2+3+...+2014]

còn lại là tịt

Câu hỏi của Phan Nguyễn Hà Linh - Toán lớp 6 - Học toán với OnlineMath

Ta có:

\(S=\dfrac{1}{2}(1+2)+\dfrac{1}{3}(1+2+3)+\cdots+\dfrac{1}{2016}(1+2+\cdots+2016)\)

Áp dụng công thức:

\(1+2+\cdots+n=\dfrac{n(n+1)}{2}\)

Ta được:

\(S=\dfrac{1}{2}\times\dfrac{2\times3}{2}+\dfrac{1}{3}\times\dfrac{3\times4}{2}+\cdots+\dfrac{1}{2016}\times\dfrac{2016\times2017}{2}\)

Rút gọn từng số hạng:

\(S=\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+\cdots+\dfrac{2017}{2}\)

\(S=\dfrac{3+4+5+\cdots+2017}{2}\)

Tính tổng:

\(3+4+\cdots+2017=\dfrac{(3+2017)\times2015}{2}\)

\(=\dfrac{2020\times2015}{2}=2\,035\,150\)

Vậy:

\(S=2\,035\,150:2=1\,017\,575\)

Ta có:

\(S=\dfrac{1}{2}(1+2)+\dfrac{1}{3}(1+2+3)+\cdots+\dfrac{1}{2016}(1+2+\cdots+2016)\)

Áp dụng công thức tổng:

\(1+2+\cdots+n=\dfrac{n(n+1)}{2}\)

Ta được:

\(S=\dfrac{1}{2}\times\dfrac{2\times3}{2}+\dfrac{1}{3}\times\dfrac{3\times4}{2}+\cdots+\dfrac{1}{2016}\times\dfrac{2016\times2017}{2}\)

Rút gọn:

\(S=\dfrac{3}{2}+\dfrac{4}{2}+\cdots+\dfrac{2017}{2}\)

\(S=\dfrac{3+4+5+\cdots+2017}{2}\)

Ta có:

\(3+4+\cdots+2017=\dfrac{(3+2017)\times2015}{2}\)

\(=\dfrac{2020\times2015}{2}=2\,035\,150\)

Vậy:

\(S=\dfrac{2\,035\,150}{2}=1\,017\,575\)

a) \(S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}\)

\(\Rightarrow2S=1+\frac{1}{2}+...+\frac{1}{2^{2013}}\)

\(\Rightarrow2S-S=1-\frac{1}{2^{2014}}\)

b) Ta có : \(S=1-\frac{1}{2^{2013}}< 1\left(ĐPCM\right)\)

\(S=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2014.2015}\)

\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\)

\(S=1-\frac{1}{2015}=\frac{2014}{2015}\)

Vậy:...

\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)

\(=1-\frac{1}{2015}\)

\(=\frac{2015}{2015}-\frac{1}{2015}\)

\(=\frac{2014}{2015}\)

R.2=2+2^2+2^3+...+2^2015

R=(2+2^2+2^3+...+2^2015-1)-(1+2^2+2^3+...+2^2014)

R=(2^2015)-1

S=(2^2015)-1 / 1-(2^2015)

S=-1

bieu thuc do goi la R nhe

\(S=-2\)

\(\cdot DuyNam\)

\(S=\left(-1\right)+\left(-1\right)^2+\left(-1\right)^3+...+\left(-1\right)^{2014}+\left(-1\right)^{2015}\)

\(S=\left(-1\right)+1+\left(-1\right)+...+1+\left(-1\right)\) (2015 thừa số)

`-> S= (-1)`

các bạn ơi, giải nhanh nhanh cho mình nha. cảm ơn nhiều

(1-1/1+2)(1-1/1+2+3).....(1-1/1+2+3+....+2014)

= 1 + 2 / (2x3) + 2 : (3x4) + ... x 2 : (2014 : 2015)

= 1 + 2 x [ 1/(2x3) + 1/(3x4) + ... 1/(2014 x 2015)

= 1 + 2 x (1/2 - 1/3 + 1/3 - 1/4 - ... 1/2014 + 1/2014 - 1/2015 )

= 1 + 2 x ( 1/2 - 1/2015 )

= 1 + 1 - 2/2015

= 2 - 2/2015 

= 4028/2015