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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:
\(S=2\,035\,150:2=1\,017\,575\)
a, s1 có 2015 hạng tử
=> s1= (2014:2).-1+2015=1007.(-1)+2015=1008
Lời giải:
a,S1=1+(-2)+3+(-4)+...+(-2014)+2015
=(1-2)+(3-4)+...+(2013-2014)+2015
=-1+(-1)+...+(-1)+2015
=-1.1007+2015
=(-1007)+2015
=1008
b,S2=(-2)+4+(-6)+8+...+(-2014)+2016
=(-2+4)+(-6+8)+...+(-2014+2016)
=2+2+...+2
=2.504
=1008
c,S3=1+(-3)+5+(-7)+...+2013+(-2015)
=(1-3)+(5-7)+...+(2013-2015)
=(-2)+(-2)+...+(-2)
=(-2).504
=-1008
d,S4=(-2015)+(-2014)+(-2013)+...+2015+2016
=(-2015+2015)+...+0+2016
=0+...+0+2016
=2016
STUDY WELL !
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2015.2016}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\)
\(S=1-\frac{1}{2016}=\frac{2015}{2016}\)
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-........+\frac{1}{2015}-\frac{1}{2016}\)
\(S=\frac{1}{1}-\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+......+\left(-\frac{1}{2015}+\frac{1}{2015}\right)-\frac{1}{2016}\)
\(S=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)
S = 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/2014x2015 + 1/2015x2016
S = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2014 - 1/2015 + 1/2015 - 1/2016
S = 1 - 1/2016
S = 2015
\(A=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+....+\left[2013+\left(-2014\right)+2015\right]\)
\(A=\left(-1\right)+\left(-1\right)+....+\left(-1\right)+2015\left(\text{1007 số hạng }\left(-1\right)\right)=1008\)
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:
\(S=\dfrac{2\,035\,150}{2}=1\,017\,575\)