tính tổng
a)S1= 1 +(-2)+3+(-4)+......+2017+(-2018)
H24
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Tính nhanh
a) S1= 1+(-2)+3+(-4)+.......+2016+(-2017)
b) S2 = 2-4+6-8+..........+2016-2018
s1=1-2+3-4+...+2017-2018
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S=1-2+3-4+...+2017-2018
S=(1-2)+(3-4)+...+(2017-2018)
S=(-1)+(-1)+...+(-1)
S=(-1).1009
S= - 1009
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+ Xét số số hạng của tổng trên là: (2018 - 1) : 1 + 1 = 2018 (số hạng)
= 2018 : 2 = 1009 (cặp)
S1 = 1 - 2 + 3 - 4 + ... + 2017 - 2018
= (1 - 2) + (3 - 4) + ... + (2017 - 2018) } 1009 cặp
= (-1) + (-1) + ... + (-1) } 1009 số (-1)
= (-1) . 1009
= -1009Vậy tổng S1 = -1009
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tính tổng
S1=1+(-2)+3+(-4)+......+2017+(-2018)
S2=1+(-3)+5+(-7)+...+(-1999)+2001
S3=13-12+11+10-9+8-7-6+5-4+3+2-1
viết cả lời giải nhé
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/1)+(2019/2)+(2019/3)+(2019/4)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/2)+(2019/3)+(2019/4)+(2019/5)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
Tính tỉ số A/B biết:
A=1/2 + 1/3 + 1/4 + ... + 1/2017 + 1/2018 + 1/2019
B=2018/1 + 2017/2 + 2016/3 + ... + 2/2017 + 1/2018
\( S =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1} {2019}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right) \)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(\(\Rightarrow S=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\) \(\Rightarrow S=P\)\)
\(B=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{1}{2018}\)
\(B=1+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{1}{2018}+1\right)\)
\(B=\frac{2019}{2019}+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2018}\)
\(B=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)\)
ta có \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}=\frac{1}{2019}\)
Cho A=1/2018+2/2017+3/2016+...+2017/2+2018
B=1/2+1/3+1/4+....+1/2019
Tính A/B
\(A=\frac{1}{2018}+\frac{2}{2017}+...+\frac{2017}{2}+2018\)
\(=\left(\frac{1}{2018}+1\right)+\left(1+\frac{2}{2017}\right)+...+\left(\frac{2017}{2}+1\right)+1\)(2018 số hạng 1)
\(=\frac{2019}{2018}+\frac{2019}{2017}+...+\frac{2019}{2}+\frac{2019}{2019}=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)\)
Mà \(B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)
=> Khi đó : \(\frac{A}{B}=\frac{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}=2019\)
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S1= 1+(-3)+5+(-7)+...+2017
S2= (-2)+4+(-6)+8+...+2016
Tính S1+S2
S1=1613 S2=404
S1+S2=2017
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S1= 1+(-3)+5+(-7)+...+2017
S2= (-2)+4+(-6)+8+...+2016
Tính S1+S2
Giải:S1+S2=1-2-3+4+5-6-7+8+...........+2013-2014-2015+2016+2017
=0+0+...........+0+2017
=2017
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tính tổng
a, 1+(-2)+3+(-4)+........+99+(-100)+101
A=1-2+3-4+5-6+.....+99-100+101
A = (1 - 2 ) + ( 3 - 4 ) + ( 5 - 6 ) + ... + ( 99 - 100 ) + 101
A = ( -1 ) + ( -1 ) + ( -1 ) + ... + ( -1 ) + 101
A = ( -1 ) . 50 + 101
A = -50 + 101
A = 51
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