\(A=1+5+5^2+5^3+...+5^{100}\)

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

\(5A-A=5+5^2+5^3+...+5^{100}+5^{101}-\left(1+5+5^2+5^3+...+5^{100}\right)\)

\(4A=5^{101}-1\)

\(A=\frac{5^{101}-1}{4}\)

A = 1+5+5²+5³+...+5¹⁰⁰

5A = 5+5²+5³+5⁴+....+5¹⁰¹

4A = 5A - A = 5¹⁰¹ - 1

=> A = \(\frac{5^{101}-1}{4}\)

thank you

A=1+2^2+...+2^100

2A=2+2^2+2^3+...+2^101

2A=2^101-1

A=(2^101-1):2

\(B=5^1+5^2+...+5^{199}\)

\(\Rightarrow5B=5^2+5^3+...+5^{200}\)

\(\Rightarrow5B-B=\left(5^2+5^3+...+5^{200}\right)-\left(5^1+5^2+...+5^{199}\right)\)

\(\Rightarrow4B=5^{200}-5\)

\(\Rightarrow B=\frac{5^{200}-5}{4}\)

A=1+5+5²+...+5¹⁰⁰

=>5A=5+5²+5³+...+5¹⁰¹

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A=1+5+5²+....5¹⁰⁰

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4A=5¹⁰¹-1

=>A=5¹⁰¹-1/4

5A=5+5^2+5^3+.....+5^101

5A-A=5^101-1

A=\(\frac{5^{101}-1}{4}\)

= 1 + 5^{1+2+3+...+100}

= 1 + 5⁵⁰⁵⁰.

đặt A=  1+5+5^2+...+5^100

     5A=5x(1+5+5^2+...+5^100)

5A= 5+5^2+...+5^100+5^101

5A-A= (5+5^2+...+5^100+5^101)-(  1+5+5^2+...+5^100)

4A=5^101-1

A=\(\frac{5^{101-1}}{4}\)

Ta có:

\(K=1+5^2+5^3+...+5^{100}\)

\(\Rightarrow5K=5+5^3+5^4+...+5^{101}\)

\(\Rightarrow5K-K=5+5^3+5^4+...+5^{101}-1-5^2-5^3-...-5^{100}\)

\(\Rightarrow4K=5^{101}-4\)

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

Ta có K = 1 + 5² + 5⁴ + 5⁶ + ... + 5¹⁰⁰

=> 5².K = 25K =  5² + 5⁴ + 5⁶ + 5⁸ + ... + 5¹⁰²

Khi đó 25K - K = (5² + 5⁴ + 5⁶ + 5⁸ + ... + 5¹⁰²) - (1 + 5² + 5⁴ + 5⁶ + ... + 5¹⁰⁰)

           => 24K = 5¹⁰² - 1

           =>     K = \(\frac{5^{102}-1}{24}\)

Vậy K = \(\frac{5^{102}-1}{24}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

A=20+21+22+23+...++23+...+250250

2�=2+22+23+...+2512A=2+22+23+...+251

2�−�=�=251−202A−A=A=251−20

�=5+52+53+...+599+5100B=5+52+53+...+599+5100

5�=52+53+54+...+5100+51015B=52+53+54+...+5100+5101

5�−�=4�=5101−55B−B=4B=5101−5

�=5101−54B=45101−5​

�=3−32+33−34+...+C=3−32+33−34+...+32007−32008+32009−3201032007−32008+32009−32010

3�=32−33+34−35+...−32008+32009−32010+320113C=32−33+34−35+...−32008+32009−32010+32011

3�+�=4�=32011+33C+C=4C=32011+3

�=32011+34C=432011+3​

�100=5+5×9+5×92+5×93+...+5×999S100​=5+5×9+5×92+5×93+...+5×999

�100=5×(1+9+92+93+...+999)S100​=5×(1+9+92+93+...+999)

9�100=5×(9+92+93+...+999+9100)9S100​=5×(9+92+93+...+999+9100)

9�100−�100=8�100=5×(9100−1)9S100​−S100​=8S100​=5×(9100−1)

�100=5×(9100−1)8S100​=85×(9100−1)​

\(A=1+2+2^2+...+2^{51}\)

\(2A=2+2^2+2^3+...+2^{52}\)

\(2A-A=\left(2+2^2+2^3+...+2^{52}\right)-\left(1+2+2^2+...+2^{51}\right)\)

\(A=2^{52}-1\)

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

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

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

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

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