Tinh gia tri biểu thúc:A =6 +5^2+5^3+5^4+...+5^1996+5^1997
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A=6+52+53+54+...+51996+51997
A = 1 + 5 + 52 + 53 + ... + 51996 + 51997
5A = 5 + 52 + 53 + 54 + ... + 51997 + 51998
5A - A = ( 5 + 52 + 53 + 54 + ... + 51997 + 51998 ) - ( 1 + 5 + 52 + 53 + ... + 51996 + 51997 )
4A = 51998 - 1
\(\Rightarrow A=\frac{5^{1998}-1}{4}\)
\(A=6+5^2+5^3+...+5^{1996}+5^{1997}\\ A=1+5+5^2+5^3+...+5^{1996}+5^{1997}\)
\(5A=5+5^2+5^3+...+5^{1996}+5^{1998}\)
\(5A-A=\left(5+5^2+5^3+...+5^{1996}+5^{1998}\right)-\left(1+5+5^2+5^3+...+5^{1996}+5^{1997}\right)\)
\(4A=5^{1998}-1\\ A=\frac{5^{1998}-1}{4}\)
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a) A=1-2-3+4+5-6-7+.....+1996+1997-1998-1999+2000
=(1-2-3+4)+(5-6-7+8)+...+(1997-1998-1999+2000)
=0
b) B=1-3+5-7+....+2001-2003+2005
=(1-3)+(5-7)+...+(2001-2003)+2005
=-2.501+2005
=-1002+2005
=1003
c) C=1-2-3+4+5-6-7+8+.....+1993-1994-1995+1996+1997
=(1-2-3+4)+(5-6-7+8)+...+(1993-1994-1995+1996)+1997
=1997
d) D=1000+998+996+......+10-999-997-995-...-11
=(1000-999)+(998-997)+(996-995)+....+(12-11)+10
=1.495+10
=595
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Ta có: \(A=6+5^2+5^3+5^4+...+5^{1996}+5^{1997}=1+5+5^2+5^3+...+1^{1997}\)
\(\Rightarrow5A=5+5^2+5^3+5^4+...+5^{1997}+5^{1998}\)
\(\Rightarrow5A-A=\left(5+5^2+5^3+5^4+...+5^{1997}+5^{1998}\right)-\left(1+5+5^2+5^3+...+5^{1996}+5^{1997}\right)\)
\(\Rightarrow4A=5^{1998}-1\Rightarrow A=\dfrac{5^{1998}-1}{4}\)
Vậy ...
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mình cũng đang làm câu đó nghĩ ra rồi nhưng ko biết có đúng ko
đặt 5^2+...+5^1997
=5A-A=5^3+...+5^1998 - 5^2-...- 5^1997
= 4A= 5^1998-5^2
=5^1998-5^2/4