A = 2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - ...-2-1

=> 2A = 2¹⁰¹ - 2¹⁰⁰ - 2⁹⁹-...-2² - 2

=> 2A-A = 2¹⁰¹ - 2¹⁰⁰ - 2¹⁰⁰ + 1

A = 2¹⁰¹ - 2¹⁰⁰.(1+1) + 1

A = 2¹⁰¹ - 2¹⁰⁰. 2+1

A = 2¹⁰¹- 2¹⁰¹+1

A = 1

b) B = 1 - 5 + 5² - 5³+...+5⁹⁸-5⁹⁹

=> 5B = 5 - 5²+5³-5⁴+...+5⁹⁹-5¹⁰⁰

=> 5B+B = -5¹⁰⁰+1

6B = -5¹⁰⁰+1

\(B=\frac{-5^{100}+1}{6}\)

Vô đây:hoi-dap/question/529337.html

   A=3 + 3^2 + 3^3 +...+3^99

3.A=      3^2 + 3^3 +...+3^99 + 3^100

2.A=      3^100 - 3

   A=     (3^100 - 3) : 2

làm vậy có đúng ko bn?

Ta có:

5A= 5^2-5^3+5^-5^5+...-5^99+5^100

=>6A=5+5^100

=>A= (5+5^100):6

\(A=1-5+5^2-5^3+...+5^{98}-5^{99}\)

\(\Rightarrow5A=5-5^2+5^3-5^4+...+5^{99}-5^{100}\)

\(\Rightarrow5A+A=\left(5-5^2+5^3-5^4+...+5^{99}-5^{100}\right)+\left(1-5+5^2-5^3+...+5^{98}-5^{99}\right)\)\(\Rightarrow6A=1-5^{100}\)

\(\Rightarrow A=\dfrac{1-5^{100}}{6}\)

Ta có:

\(A=1-5+5^2-5^3+.....+5^{98}-5^{99}\)

\(\Rightarrow5A=5-5^2+5^3-5^4+......+5^{99}-5^{100}\)

\(\Rightarrow5A+A=\left(5-5^2+5^3-5^4+.......+5^{99}-5^{100}\right)+\left(1-5+5^2-5^3+......+5^{98}-5^{99}\right)\)

\(\Rightarrow6A=-5^{100}+1\)

\(\Rightarrow A=\dfrac{1-5^{100}}{6}\)

Chúc bạn học tốt!!!

Đề 1 nhé: Ta có: B= 1 +5 +5^2 +...+5^97 + 5^98 +5^99 (1)

5B = 5 + 5^2 + 5^3 +...+5^98 +5^99 + 5^100 (2)

Trừ vế với vế của (2) cho (1) ta có:

4B = 5^100 - 1

=>B = (5^100 - 1)/4

Tk nha bn!

Đề 2 tương tự thôi.

\(B=1+5+5^2+5^3+...+5^{98}+5^{99}\)

\(\Rightarrow5B=5+5^2+5^3+....+5^{98}+5^{99}+5^{100}\)

\(\Rightarrow5B-B=\left(5+5^2+5^3+....+5^{100}\right)-\left(1+5+5^2+...+5^{99}\right)\)

\(\Rightarrow4B=5^{100}-1\)

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

(CÒn lại tương tự: ĐS: \(\frac{5^{99}-1}{4}\) )

5A=5²-5³+5⁴-.....+5⁹⁸-5⁹⁹-5¹⁰⁰

^5A+A=(5²-5³+5⁴-.....+5⁹⁸-5⁹⁹-5¹⁰⁰)+(5-5²+5³-5⁴+.....-5⁹⁸+5⁹⁹)

6A=-5¹⁰⁰+5

A(-5¹⁰⁰+5):6