A= 1+5+5²+5³+5⁴+...+5¹⁹⁹⁶+5¹⁹⁹⁷=\(\left(5-1\right).\left(1+5+5^2+5^3+...+5^{1997}\right).\frac{1}{4}\)

                                                    \(=\left(5^{1998}-1\right)\frac{1}{4}\)

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

\(\Rightarrow A=\frac{5^{1998}-1}{4}\)

\(A=6+5^2+5^3+...+5^{1996}+5^{1997}\\ A=1+5+5^2+5^3+...+5^{1996}+5^{1997}\)

\(5A=5+5^2+5^3+...+5^{1996}+5^{1998}\)

\(5A-A=\left(5+5^2+5^3+...+5^{1996}+5^{1998}\right)-\left(1+5+5^2+5^3+...+5^{1996}+5^{1997}\right)\)

\(4A=5^{1998}-1\\ A=\frac{5^{1998}-1}{4}\)

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

Ta có: \(A=6+5^2+5^3+5^4+...+5^{1996}+5^{1997}=1+5+5^2+5^3+...+1^{1997}\)

\(\Rightarrow5A=5+5^2+5^3+5^4+...+5^{1997}+5^{1998}\)

\(\Rightarrow5A-A=\left(5+5^2+5^3+5^4+...+5^{1997}+5^{1998}\right)-\left(1+5+5^2+5^3+...+5^{1996}+5^{1997}\right)\)

\(\Rightarrow4A=5^{1998}-1\Rightarrow A=\dfrac{5^{1998}-1}{4}\)

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