A = 1 +  3  + 3² + 3³ + ... + 3¹⁰⁰

3A = 3 + 3² + 3³ +3⁴+ .... + 3¹⁰¹

3A - A = (3 + 3² + 3⁴ + ... + 3¹⁰¹) - (1 + 3 + 3² + 3³ + ... + 3¹⁰⁰)

2A     = 3 + 3² + 3⁴ + ... + 3¹⁰¹ - 1 - 3 - 3² - 3³ - ... - 3¹⁰⁰

2A = (3 - 3) + (3² - 3²) + ... + (3¹⁰⁰ - 3¹⁰⁰) + (3¹⁰¹ - 1)

2A = 3¹⁰¹ - 1

A = \(\dfrac{3^{101}-1}{2}\)

\(A=3^{100}-3^{99}+3^{98}-...-3+1\\ \Rightarrow\dfrac{1}{3}A=3^{99}-3^{98}+3^{97}-...-1+\dfrac{1}{3}\\ \Rightarrow\dfrac{4}{3}A=3^{100}+\dfrac{1}{3}\\ \Rightarrow A=\dfrac{3^{101}}{4}+\dfrac{1}{4}\)

\(B=1+3+3^2+3^3+...+3^{100}+3^{101}\)

\(\Rightarrow3B=3+3^2+3^3+3^4+...+3^{101}+3^{102}\)

\(\Rightarrow3B-B=3^{102}-1\)

\(\Leftrightarrow2B=3^{102}-1\)

\(\Leftrightarrow B=\dfrac{3^{102}-1}{2}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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)​

2A = 3A - A = (3 + 3² + 3³ + ... + 3¹⁰¹) - (1 + 3 + 3² + 3³ + ... + 3¹⁰⁰)

2A = 3¹⁰¹ - 1

A = \(\frac{3^{101}-1}{2}\)

3B = 4B - B = (4 + 4² + ... + 4⁵¹) - (1 + 4 + 4² + ... + 4⁵⁰)

3B = 4⁵¹ - 1

B = \(\frac{4^{51}-1}{3}\)

2A = 3A - A = ( 3 + 3^{2  +}  3³  +  ...  +  3¹⁰¹ )  - ( 1 + 3 + 3² +  3³  +  ... + 3¹⁰⁰ ) 

2A = 3¹⁰¹ - 1

A =\(3^{101}-1\): 2

3B =  4B - B = ( 4 + 4²  + ... +  4⁵¹) - ( 1 + 4 + 4² +...+ 4⁵⁰ )

3B = 4⁵¹ - 1

B = 4⁵¹ - 1 : 3

A=1+3+3²+...+3¹⁰⁰

3A=3+3²+3³+...+3¹⁰¹

=>3A+1=1+3+3²+...+3¹⁰⁰+3¹⁰¹=A+3¹⁰¹

=>3A-A=3¹⁰¹-1

2A=3¹⁰¹-1

A=(3¹⁰¹-1)/2

B=1+4+4²+...+4⁵⁰

4B=4+4²+...+4⁵¹

4B+1=1+4+4²+...+4⁵⁰+4⁵¹=B+4⁵¹

=>4B-B=4⁵¹-1

3B=4⁵¹-1

B=(4⁵¹-1)/3

Đầu luỹ thừa đuôi số hạng 

A=\(\frac{3^{101}-1}{2}\)

B=\(\frac{4^{51}-1}{3}\)

a: Tổng các số hạng là:

\(\dfrac{\left(220+1\right)\cdot220}{2}=24310\)

Ta có: A+1=2x

\(\Leftrightarrow2x=24311\)

hay \(x=\dfrac{24311}{2}\)

\(D=3^{100}+3^{101}+...+3^{149}+3^{150}\)

nên \(3D=3^{101}+3^{102}+...+3^{150}+3^{151}\)

\(\Leftrightarrow2\cdot D=3^{151}-3^{100}\)

hay \(D=\dfrac{3^{151}-3^{100}}{2}\)

\(3D=3^{101}+3^{102}+3^{103}+...+3^{150}+3^{151}\\ 3D-D=3^{151}-3^{100}\\ 2D=3^{151}-3^{100}\\ D=\dfrac{3^{151}-3^{100}}{2}\)

a, A = 3¹ + 3² + 3³ + 3⁴ +...+ 3⁹⁹ + 3¹⁰⁰

3A = 3(3¹ + 3² + 3³ + 3⁴ +...+ 3⁹⁹ + 3¹⁰⁰)

3A = 3² + 3³ + 3⁴ + 3⁵ +...+ 3¹⁰⁰ + 3¹⁰¹

3A - A = (3² + 3³ + 3⁴ + 3⁵ +...+ 3¹⁰⁰ + 3¹⁰¹) - (3¹ + 3² + 3³ + 3⁴ +...+ 3⁹⁹ + 3¹⁰⁰)

2A = 3¹⁰¹ - 3¹ = 3¹⁰¹ - 3

A = \(\frac{3^{101}-3}{2}\)

b, A = 3¹ + 3² + 3³ + 3⁴ +...+ 3⁹⁹ + 3¹⁰⁰

A = (3¹ + 3² + 3³ + 3⁴) +...+ (3⁹⁷ + 3⁹⁸ + 3⁹⁹ + 3¹⁰⁰)

A = (3¹ + 3² + 3³ + 3⁴⁾) +...+ 3⁹⁶(3¹ + 3² + 3³ + 3⁴)

A = 120 +...+ 3⁹⁶.120

A = 120(1 +...+ 3⁹⁶) chia hết cho 40 (ĐPCM)