a) 2 +4+6+8+...+2018

= ( 2018+2) x 1009 : 2

= 2020 x 1009 : 2

= 1009 x (2020:2)

= 1009 x 1010

= 1 019 090

b) S = 10 + 10² + 10³ + ...+ 10¹⁰⁰

=> 10.S = 10² + 10³ + 10⁴ +...+ 10¹⁰¹

=> 10.S - S = 10¹⁰¹-10

9.S=10¹⁰¹- 10

\(\Rightarrow S=\frac{10^{101}-10}{9}\)

c) \(S=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)

\(\Rightarrow5S=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)

\(5S-S=1-\frac{1}{5^{100}}\)

\(4S=1-\frac{1}{5^{100}}\)

\(S=\frac{1-\frac{1}{5^{100}}}{4}\)

e cx ko nx, e ms hok lp 7 thoy, sang hè ms lp 8! e sr cj nhiều nha!

d) \(S=\frac{1!}{3!}+\frac{2!}{4!}+\frac{3!}{5!}+...+\frac{2018!}{2020!}\)

\(S=\frac{1}{1.2.3}+\frac{1.2}{1.2.3.4}+\frac{1.2.3}{1.2.3.4.5}+...+\frac{1.2.3...2018}{1.2.3...2020}\)

\(S=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2019.2020}\)

\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(S=\frac{1}{2}-\frac{1}{2020}\)

\(S=\frac{1009}{2020}\)

a;b;c có những câu tương tự rồi, ko giải lại nx

d) \(S=\frac{1!}{3!}+\frac{2!}{4!}+...+\frac{2018!}{2020!}\)

\(S=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2019.2020}\)

\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(S=\frac{1}{2}-\frac{1}{2020}\)

b tự làm nốt nha

# Tính lũy thừa
a = 3
n = 2
power = a ** n
print(power)  # Kết quả: 9

# Tính giai thừa
n = 5
factorial = 1
for i in range(1, n+1):
    factorial *= i
print(factorial)  # Kết quả: 120

\(4!-3!\)

\(=1.2.3.4-1.2.3\)

\(=24-6\)

\(=18\)

4!-3!=(1.2.3.4)-(1.2.3)=24-6=18

\(A=81.\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right].\frac{158158158}{711711711}\)

\(A=81.\left[\frac{12.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{158}{711}\)

\(A=81.\left(\frac{12}{4}:\frac{5}{6}\right).\frac{2}{9}\)

\(A=81.3.\frac{6}{5}.\frac{2}{9}\)

\(A=\frac{324}{5}\)

Nhớ là: THANKS YOU VERY "MUCH" chứ không phải là  THANKS YOU VERY "MATH"!!!

\(A=81.\frac{158158158}{711711711}.\frac{12.\left(\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5.\left(\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6.\left(\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\)

\(=81.\frac{158}{711}.\frac{12}{4}:\frac{5}{6}=\frac{1422}{79}.3.\frac{6}{5}=\frac{1422.3.6}{79.5}=\frac{25596}{395}\)

\(=\frac{324}{5}\)

Ta có: \(D=\dfrac{\left(n+1\right)!}{\left(n-1\right)!}-n\)

\(=\dfrac{1\cdot2\cdot...\cdot\left(n-1\right)\cdot n\cdot\left(n+1\right)}{1\cdot2\cdot...\cdot\left(n-1\right)}-n\)

\(=n\left(n+1\right)-n\)

\(=n^2\)