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a) \(A=2+2^2+2^3+................+2^{100}\)
\(\Rightarrow2A=2^2+2^3+2^4+................+2^{100}+2^{101}\)
\(\Rightarrow2A-A=\left(2^2+2^3+..............+2^{100}+2^{101}\right)-\left(2+2^2+............+2^{100}\right)\)
\(\Rightarrow A=2^{101}-2\)
b) \(B=1+3+3^2+..................+3^{2009}\)
\(\Rightarrow3B=3+3^2+3^3+..................+3^{2009}+3^{2010}\)
\(\Rightarrow3B-B=\left(3+3^2+...............+3^{2010}\right)-\left(1+3+3^2+.............+3^{2009}\right)\)
\(\Rightarrow2B=3^{2010}-1\)
\(\Rightarrow B=\dfrac{3^{2010}-1}{2}\)
c) \(C=4+4^2+4^3+................+4^n\)
\(\Rightarrow4C=4^2+4^3+.................+4^n+4^{n+1}\)
\(\Rightarrow4C-C=\left(4^2+4^3+.............+4^n+4^{n+1}\right)-\left(4+4^2+............+4^n\right)\)
\(\Rightarrow3C=4^{n+1}-4\)
\(\Rightarrow C=\dfrac{4^{n+1}-4}{3}\)
(1+2+3+.....+100).(12+22+32+.....+102).(65.111-13.15.37)
=(1+2+3+.....+100).(12+22+32+.....+102)(65.111-13.555)
=(1+2+3+.....+100).(12+22+32+.....+102)(65.111-13.5.111)
=(1+2+3+.....+100).(12+22+32+.....+102)[111(65-65)]
=(1+2+3+.....+100).(12+22+32+.....+102)(100.0)
=(1+2+3+.....+100).(12+22+32+.....+102)0
=0
Ta có : (1+2+3+.....+100).(12+ 22 +.....+102 ).(65.111 - 13.15.37)
=A .B.(65.111-13.5.3.37)
=A.B.(65.111-65.111)
=A.B.0
=0
Tính nhanh:
a) (1+2+3+..+100) x (12 +22+32+...+...102) x ( 65.111-13.15.37)
b)19x64+76x34
c)12x35+65x13
a Ta có
(1+2+3....+100).(1^2+2^2+..+10^2).(13.5.3.37-13.15.37)
=(1+2...+100).(1^2+2^2+...+10^2).0=0
b Ta co
19.64+76.34=19.4.16+76.34=76.16+76.34=76(16+34)=76.50=3100 ( cau c tuong tu )
\(D=\left(2^9.3+2^9.5\right)-2^{12}\)
\(D=2^9.\left(3+2\right)-2^{12}\)
\(D=2^9.5-2^{12}\)
\(D=512.5-4096\)
\(D=2560-4096\)
\(D=-1536\)
\(\left(1+2+3+...+100\right).\left(1^2+2^2+3^2+...+100^2\right).\left(65.111-13.15.37\right)\)
\(=\left(1+2+3+...+100\right).\left(1^2+2^2+3^2+...+100^2\right).\left(7215-7215\right)\)
\(=\left(1+2+3+...+100\right).\left(1^2+2^2+3^2+...+100^2\right).0\)
\(=0\)
D=(29.3+29.5)-212
D=((29.(3+5))-212
D=(29.8)-212
D=(29.23)-212
D=29+3-212
D=212-212
D=0
(1+2+3+...+100).(12+22+32+....+1002).(65.111-13.15.37)
=(1+2+3+...+100).(12+22+32+....+1002).7215-7215
=(1+2+3+...+100).(12+22+32+....+1002).0
=0
C=210-2
C=29+1-2
C=29.2-2
C=2.(29-1)
C=2.(512-1)\
C=2.511
C=1022
F=1+31+32+33+......+3100
F=3+31+32+33+......+3100
3F=3.(3+31+32+33+......+3100)
3F=32+32+33+34+......+3100
3F-F=3100+32-3-3
2F=3100+9-3-3
F=\(\frac{3^{100}+3}{2}\)
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