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c, 4C= (1.2.3+2.3.4+3.4.5+...+8.9.10) .4
==> 4C= [1.2.3.(4-0) + 2.3.4-(5-1) + 8.9.10.(11-7)
==>4C= 1.2.3.4 - 1.2.3.4+ 2.3.4.5-2.3.4.5 + 7.8.9.10- 7.8.9.10 + 8.9.10.11
==> 4C= 8.9.10.11=7920
==> C= 7920 :4=1980
a, Ta có: 3A= 1.2.3+2.3.3+3.4.3+...+99.100.3
3A=1.2.(3-0) + 2.3.(4-1)+ 3.4.(5-2)+ ... + 99.100.( 101-98)
3A=(1.2.3 + 2.3.4+ 3.4.5+ 99.100.101) - (0.1.2 +1.2.3+ 2.3.4 + ... + 98.99.100)
3A= 99.100.101 - 0.1.2
3A= 999900 - 0
3A= 999900
==> A= 999900 : 3
==> A= 333300
Bài 2:
a) Ta có: \(f\left(2\right)-f\left(-1\right)=7\)
\(\Leftrightarrow2\left(m-1\right)-\left(-1\right)\cdot\left(m-1\right)=7\)
\(\Leftrightarrow2m-2+m-1=7\)
\(\Leftrightarrow3m-3=7\)
\(\Leftrightarrow3m=10\)
hay \(m=\dfrac{10}{3}\)
5,Ta có
A=1/2+1/2^2+1/2^3+...+1/2^100
2A=1+1/2+1/2^2+1^2/3+...+1/2^99
2A-A=(1+1/2+1/2^2+1^2/3+...+1/2^99)-(1/2+1/2^2+1/2^3+...+1/2^100)
A=1-1/2^100
Ta có : A = 1.2.3 + 2.3.4 + 4.5.6 + ..... + 98.99.100
=> 6A = 1.2.3.4 - 1.2.3.4 + 2.3.4.5 - 2.3.4.5 + ...... + 98.99.100.101
=> 6A = 98.99.100.101
=> A = \(\frac{98.99.100.101}{6}=16331700\)
có 20172 đồng dư 1 mod (3)
=> (20172)50 đồng dư 1 mod (3)
=> (20172)50-1 đồng dư 1-1 = 0 mod (3)
=> dpcm
Câu a)
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(=\left(2^{100}+2^{99}+2^{98}+2^{97}+...+2^2+2\right)-2\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)\)
\(=\left(2^{100}+2^{99}+2^{98}+2^{97}+...+2^2+2\right)-\left(2^{100}+2^{98}+2^{96}+...+2^4+2^2\right)\)
\(=2^{99}+2^{97}+2^{95}+...+2^3+2\)
\(=\frac{2^2\cdot\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)-\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)}{3}\)
\(=\frac{\left(2^{101}+2^{99}+2^{97}+...+2^5+2^3\right)-\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)}{3}\)
\(=\frac{2^{101}-2}{3}\)
\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2015.2016.2017}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{2.4}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\)
\(2B=\frac{1}{1.2}-\frac{1}{2016.2017}\)
\(B=\frac{\frac{1}{1.2}-\frac{1}{2016.1017}}{2}\)
Bài 2 :
\(B=2014\cdot2020\)
\(B=\left(2017-3\right)\left(2017+3\right)\)
\(B=2017^2-3^2\)
\(B=2017^2-9< A=2017^2\)
Vậy \(B< A\)
\(B=2014.2020\)
\(B=\left(2017-3\right)\left(2017+3\right)\)
\(B=\left(2017-3\right).2017+\left(2017+3\right).3\)
\(B=2017^2-3.2017+2017.3+3^2\)
\(B=2017^2-3^2< 2017^2=A\)
Vậy A > B
_Hok tốt_
!!!