\(A=\left(1+5^2+5^4+...+5^{200}\right)\)

=>\(5^2.A=5^2.\left(1+5^2+5^4+...+5^{200}\right)\)

=>\(25A=5^2+5^4+5^8+...+5^{202}\)

=>\(25A-A=\left(5^2+5^4+5^6+...+5^{202}\right)-\left(1+5^2+5^4+...+5^{200}\right)\)

=>\(24A=5^{202}-1\Rightarrow A=\frac{5^{202}-1}{24}\)

1. Đặt A = 1 + 5² + 5⁴ + ... + 5^200 

Ta có: 5²A = 5² +  5⁴ +  5⁶ + ... + 5^202

25A - A = (5² + 5⁴ + ... + 5²⁰²) - (1 + 5² + ... + 5²⁰⁰)

24A = 5²⁰² - 1     =>    A = (5²⁰² - 1) : 24 

2. Ta có : 777²²² = (777²)¹¹¹

                222⁷⁷⁷= (222⁷)¹¹¹¹¹

Vì 777² < 222⁷ => (222⁷)¹¹¹ > (777²)¹¹¹ 

    =>  222⁷⁷⁷ > 777²²² 

chi lam duoc bai 2 thôi

kết qua la:(5^201-5)/4

abc +acb= bca

= (2√2 - 3√2 + 10)√2 - √5

= 2.(√2)2 - 3.(√2)2 + √10.√2 - √5

= 4 - 6 + √20 - √5 = -2 + 2√5 - √5

= -2 + √5

= 0,2.10.√3 + 2|√3 - √5|

s

= 2√3 + 2(√5 - √3)

= 2√3 + 2√5 - 2√3 = 2√5

\(A=2^0+2^1+2^2\)\(+2^3+...+\)\(2^{50}\)

\(2A=2+2^2+2^3+...+2^{51}\)

\(2A-A=A=2^{51}-2^0\)

\(B=5+5^2+5^3+...+5^{99}+5^{100}\)

\(5B=5^2+5^3+5^4+...+5^{100}+5^{101}\)

\(5B-B=4B=5^{101}-5\)

\(B=\frac{5^{101}-5}{4}\)

\(C=3-3^2+3^3-3^4+...+\)\(3^{2007}-3^{2008}+3^{2009}-3^{2010}\)

\(3C=3^2-3^3+3^4-3^5+...-3^{2008}+3^{2009}-3^{2010}+3^{2011}\)

\(3C+C=4C=3^{2011}+3\)

\(C=\frac{3^{2011}+3}{4}\)

\(S_{100}=5+5\times9+5\times9^2+5\times9^3+...+5\times9^{99}\)

\(S_{100}=5\times\left(1+9+9^2+9^3+...+9^{99}\right)\)

\(9S_{100}=5\times\left(9+9^2+9^3+...+9^{99}+9^{100}\right)\)

\(9S_{100}-S_{100}=8S_{100}=5\times\left(9^{100}-1\right)\)

\(S_{100}=\frac{5\times\left(9^{100}-1\right)}{8}\)

$A=2^0+2^1+2^2$$+2^3+...+$$2^{50}$

$2A=2+2^2+2^3+...+2^{51}$

$2A-A=A=2^{51}-2^0$

$B=5+5^2+5^3+...+5^{99}+5^{100}$

$5B=5^2+5^3+5^4+...+5^{100}+5^{101}$

$5B-B=4B=5^{101}-5$

$B=\genfrac{}{}{0ex}{}{5^{101}-5}{4}$

$C=3-3^2+3^3-3^4+...+$$3^{2007}-3^{2008}+3^{2009}-3^{2010}$

$3C=3^2-3^3+3^4-3^5+...-3^{2008}+3^{2009}-3^{2010}+3^{2011}$

$3C+C=4C=3^{2011}+3$

$C=\genfrac{}{}{0ex}{}{3^{2011}+3}{4}$

$S^{100}=5+5\times 9+5\times 9^2+5\times 9^3+...+5\times 9^{99}$

$S^{100}=5\times \left(1+9+9^2+9^3+...+9^{99}\right)$

$9S^{100}=5\times \left(9+9^2+9^3+...+9^{99}+9^{100}\right)$

$9S^{100}-S^{100}=8S^{100}=5\times \left(9^{100}-1\right)$

$S^{100}=\genfrac{}{}{0ex}{}{5\times \left(9^{100}-1\right)}{8}$

Lời giải:

$A=1+5^2+5^4+5^6+...+5^{198}+5^{200}$

$5^2A=5^2+5^4+5^6+5^8+...+5^{200}+5^{202}$

$\Rightarrow 5^2A-A=5^{202}-1$

$\Rightarrow 24A=5^{202}-1$

$\Rightarrow A=\frac{5^{202}-1}{24}$