A=1-6+11-16+21-26+...........+1991-1996+2001-2006+2010
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\(B=1-6+11-16+...+2001-2006+2010\)
\(=\left(1-6\right)+\left(11-16\right)+...+\left(2001-2006\right)+2010\)
\(=-\frac{5.\left(\frac{2006-1}{5}+1\right)}{2}+2010=-1005+2010=1005\)
a) ( - 30 ) + ( - 29 ) + ( - 28 ) + ... + 48 + 49 + 50
= [ ( - 30 ) + 30 ] + [ ( - 29 ) + 29 ] + [ ( - 28 ) + 28 ] + ..... + 48 + 49 + 50
= 0 + 0 + 0 + .... + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 + 50
= 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 + 50
= ( 31 + 49 ) + ( 32 + 48 ) + ( 33 + 47 ) + ( 34 + 46 ) + ( 35 + 45 ) + ( 36 + 44 ) + ( 37 + 43 ) + ( 38 + 42 ) + ( 39 + 41 ) + 50
= 80 + 80 + 80 + 80 + 80 + 80 + 80 + 80 + 80 + 50
= 80 . 9 + 50
= 720 + 50
= 770
A=1991x1999=
(1995-4)1999
=1995x1999-4x1999
B=1995x1995
=1995x(1999-4)
=1995x1999-1995x4>1995x1999-4x1999=A
vậy A<B
a, \(1-6+11-16+21-26+...+91-96+101\)\
\(\left(1+11+21+...+91+101\right)^{\left(1\right)}-\left(6+16+26+...+96\right)^{\left(2\right)}\)
Ta gọi (1) là B
(2) là A
Tổng dãy B là: ( 91 - 1) : 10 + 1 : 2 . ( 91 +1 ) + 101 = 561
Tổng dãy A là: ( 96 - 6) : 10 + 1 : 2 . ( 96 + 6 ) = 510
1 - 6 + 11 - 16 + 21 - 26 + ......... + 91 - 96 + 101 = 561 - 510
= 51
b, A = 1991 . 1999 = 1991 . ( 1995 + 4 ) = 1991 . 1995 + 1991 . 4
B = 1995 . 1995 = 1995 . ( 1991 + 4 ) = 1995 . 1991 + 1995 . 4
1991 < 1995 => A < B
a, Tổng trên có số số hạng là:
(31 - 1) : 2 + 1 = 16 (số)
Tổng trên là:
(31 + 1) . 16 : 2 = 256
ĐS:
b, tổng trên có số số hạng là:
(2006 - 1) : 5 + 1 = 402 (số)
Tổng trên là:
(2006 + 1) . 402 : 2 = 403407
ĐS:
a,Ta có : \(1996\equiv1\left(mod5\right)\)
\(\Rightarrow1996^{1996}\equiv1^{1996}\left(mod5\right)\)
\(1991\equiv1\left(mod5\right)\)
\(\Rightarrow1991^{1991}\equiv1^{1991}\left(mod5\right)\)
\(\Rightarrow1996^{1996}-1991^{1991}\equiv1^{1996}-1^{1991}\left(mod5\right)\)
\(\Leftrightarrow1996^{1996}-1991^{1991}\equiv0\left(mod5\right)\)
Hay \(1996^{1996}-1991^{1991}⋮5\)
b,Ta có : \(9^{1972}=\left(9^2\right)^{986}=81^{986}\)
\(7^{1972}=\left(7^4\right)^{493}=2401^{493}\)
Ta lại có : \(81\equiv1\left(mod10\right)\)
\(\Rightarrow81^{986}\equiv1^{986}\left(mod10\right)\)
\(2401\equiv1\left(mod10\right)\)
\(\Rightarrow2401^{493}\equiv1^{493}\left(mod10\right)\)
\(\Rightarrow9^{1972}-7^{1972}=81^{986}-2401^{493}\equiv1^{986}-1^{493}\left(mod10\right)\)
\(\Leftrightarrow9^{1972}-7^{1972}=81^{986}-2401^{493}\equiv0\left(mod10\right)\)
hay \(9^{1972}-7^{1972}⋮10.\)
c, Ta có : \(89\equiv1\left(mod2\right)\)
\(\Rightarrow89^{26}\equiv1^{26}\left(mod2\right)\)
\(45\equiv1\left(mod2\right)\)
\(\Rightarrow45^{21}\equiv1^{21}\left(mod2\right)\)
\(\Rightarrow89^{26}-45^{21}\equiv1^{26}-1^{21}\left(mod2\right)\)
\(\Rightarrow89^{26}-45^{21}\equiv0\left(mod2\right)\)
Hay \(89^{26}-45^{21}⋮0\)
\(1996\equiv1\left(mod5\right)\Rightarrow1996^{1996}\equiv1\left(mod5\right)\)
\(1991\equiv1\left(mod5\right)\Rightarrow1991^{1991}\equiv1\left(mod5\right)\)
\(\Rightarrow1996^{1996}-1991^{1991}\equiv1-1=0\left(mod5\right)\Leftrightarrowđpcm.\)
\(9^{1972}=\left(9^2\right)^{986}=81^{986}\equiv1\left(mod10\right)\)
\(7^{1972}=\left(7^4\right)^{493}=2401^{493}\equiv1\left(mod10\right)\)
\(\Rightarrowđpcm.\)
A=1-6+11-16+21-26+...........+1991-1996+2001-2006+2010
A=(1-6)+(11-16)+(21-26)+........+(1991-1996)+(2001-2006)+2010
A= -5+(-5)+(-5)+..........+(-5)+(-5)+2010
A=(-5).201+2010
A=-1005+2010
A=1005