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a,A=0-2+4-6+...+2010-2012
A=-2+-2+...+-2+-2 (503 cặp)
A=-2.503
A=-1006
#)Giải :
\(A=1+2+2^2+...+2^{100}\)
\(2A=2+2^2+2^3+...+2^{101}\)
\(2A-A=\left(2+2^2+2^3+...+2^{101}\right)-\left(1+2+2^2+...+2^{100}\right)\)
\(A=2^{101}-1\)
\(B=1+3^2+3^4+...+3^{100}\)
\(3^2B=3^2+3^4+3^6+...+3^{102}\)
\(3^2B-B=\left(3^2+3^4+3^6+...+3^{102}\right)-\left(1+3^2+3^4+...+3^{100}\right)\)
\(8B=3^{102}-1\)
\(B=\frac{3^{102}-1}{8}\)
\(C=1+5^3+5^6+...+5^{99}\)
\(5^2C=5^3+5^6+5^9+...+5^{102}\)
\(5^2C-C=\left(5^3+5^6+5^9...+5^{102}\right)-\left(1+5^3+5^6+...+5^{99}\right)\)
\(24C=5^{102}-1\)
\(C=\frac{5^{102}-1}{24}\)
a) A = 1 + 22 + ... + 2100
=> 2A = 22 + 23 + ... + 2101
Lấy 2A - A = (2 + 22 + ... + 2101) - (1 + 22 + ... 2100)
A = 2101 - 1
b) B = 1 + 32 + 34 + ... + 3100
=> 32B = 32 + 34 + 36 + ..... + 3102
=> 9B = 32 + 34 + 36 + ..... + 3102
Lấy 9B - B = ( 32 + 34 + 36 + ..... + 3102) - (1 + 32 + 34 + ... + 3100)
8B = 3102 - 1
B = \(\frac{3^{102}-1}{8}\)
c) C = 1 + 53 + 56 + ... + 599
=> 53.C = 53 . 56 . 59 + ... + 5102
=> 125.C = 53 . 56 . 59 + ... + 5102
Lấy 125.C - C = (53 . 56 . 59 + ... + 5102) - (1 + 53 + 56 + ... + 599)
124.C = 5102 - 1
=> C = \(\frac{5^{102}-1}{124}\)
\(\cdot DuyNam\)
\(A=-\dfrac{7}{21}+\left(1+\dfrac{1}{3}\right)\)
\(A=-\dfrac{7}{21}+\dfrac{4}{3}\)
\(A=1\)
\(B=\dfrac{2}{15}+\left(\dfrac{5}{9}+-\dfrac{6}{9}\right)\)
\(B=\dfrac{2}{15}+-\dfrac{1}{9}\)
\(B=\dfrac{1}{45}\)
\(C=\left(-\dfrac{1}{5}+\dfrac{3}{12}\right)+-\dfrac{3}{4}\)
\(C=\dfrac{1}{20}+-\dfrac{3}{4}\)
\(C=-\dfrac{7}{10}\)
a) \(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{31.34}\)
\(A=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{31}-\frac{1}{34}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{34}\right)\)
\(A=\frac{2}{3}\cdot\frac{33}{34}=\frac{11}{17}\)
b) \(B=\frac{3}{1}+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{210}\)
\(B=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{420}\) ( 3/1 = 6/2; 6/6=3/3;..)
\(B=\frac{6}{1.2}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{20.21}\)
\(B=6.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\right)\)
\(B=6.\left(1-\frac{1}{21}\right)=6\cdot\frac{20}{21}=\frac{40}{7}\)
A= [(1+101)x101:2]-(102-103)
A= 5151+1
A=5152
B= [1+(-3)]+[4+(-5)]+.......[101+(-103)]+105
B= (-2)+(-2)...........+(-2)+105
=> A>B
B=(-2)x26+105
B=(-56)+105
B= 49
cái => A>B nó nằm ở dưới cùng ấy. Nãy gõ chứ nó bị nhảy phím
A = - ( 1+2+3 +....+ 202) = - 203. 101 = -20503
B= ( 1+2-3-4) + ( 5+6-7-8) +..........+( 97+98 -99-100) + ( 101+102)
= -4 + (-4) .........+ (-4) + 203
= -4 .25 + 203 = 103