\(5A=5^2+5^3+....+5^{102}\)

\(\Rightarrow5A-A=4A=\left(5^2+5^3+....+5^{102}\right)-\left(5+5^2+.....+5^{102}\right)\)

\(\Rightarrow4A=5^{102}-5\)

\(\Rightarrow A=\frac{5^{102}-5}{4}\)

Ta có:

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

\(\Rightarrow5.A=5^2+5^3+5^4+...+5^{102}\)

\(\Rightarrow5.A-A=\left(5^2+5^3+5^4+...+5^{102}\right)-\left(5+5^2+5^3+...+5^{101}\right)\)

\(\Rightarrow4.A=5^{102}-5\)

\(\Rightarrow A=\left(5^{102}-5\right):4\)

\(A=5+5^2+5^3+5^4+....+5^{99}+5^{100}\)

\(5B=5^2+5^3+5^4+5^5+...+5^{501}\)

\(5B-B=5^2+5^3+...+5^{101}-5-5^2-5^3-....-5^{99}-5^{100}\)

\(4B=5^2+5^3+5^4+....+5^{100}+5^{101}-\left(5+5^2+5^3+...+5^{99}+5^{100}\right)\)

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

\(\Rightarrow B=\left(5^{101}-5\right):4\)

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

\(\Rightarrow5A=5^2+5^3+....+5^{102}\)

\(\Rightarrow5A-A=4A=\left(5^2+5^3+....+5^{102}\right)-\left(5+5^2+....+5^{101}\right)\)

\(\Rightarrow4A=5^{102}-5\)

\(\Rightarrow A=\frac{5^{102}-5}{4}\)

Xét biểu thức phụ : 1 (2n+3)√2n+1+(2n+1)√2n+3 = 1 √2n+1.√2n+3(√2n+1+√2n+3) = √2n+3−√2n+1 √2n+1.√2n+3[(2n+3)−(2n+1)] = √2n+3−√2n+1 2√2n+1.√2n+3 = 1 2 ( 1 √2n+1 − 1 √2n+3 )với n≥1 Áp dụng : S= 1 3√1+1√3 + 1 3√5+5√3 + 1 5√7+7√5 +...+ 1 101√103+103√101 = 1 2 ( 1 √1 − 1 √3 )+ 1 2 ( 1 √3 − 1 √5 )+ 1 2 ( 1 √5 − 1 √7 )+...+ 1 2 ( 1 √101 − 1 √103 ) = 1 2 (1− 1 √3 + 1 √3 − 1 √5 + 1 √5 − 1 √7 +...+ 1 √101 − 1 √103 ) = 1 2 (1− 1 √103 )

A=-2/3

B=1

ta có 

\(\frac{2.6.10+6.10.14+..+194.198.202}{1.3.5+3.5.7+..+97.99.101}=\frac{8.1.3.5+8.3.5.7+..+8.97.99.101}{1.3.5+3.5.7+..+97.99.101}\)

\(=\frac{8.\left(1.3.5+3.5.7+..+97.99.101\right)}{1.3.5+3.5.7+..+97.99.101}=8\)

B= 3¹¹+3¹²+3¹³+...+3¹⁰¹

=>3B= 3¹²+3¹³+3¹⁴+...+3¹⁰¹

=>3B-B= 3¹²+3¹³+3^14+...+3¹⁰¹-3¹¹ -3¹²-3¹³-...-3¹⁰¹

=>2B=3¹⁰¹-3¹¹

=>B= 2¹⁰¹-3¹¹ :2