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Gọi A = 5⁰ + 5¹ + 5² + 5³ +... + 5⁴⁹ + 5⁵⁰.
Vậy, 5A = 5¹ + 5² + 5³ +... + 5⁵⁰ + 5⁵¹.
5A - A = 4A = (5¹ + 5² + 5³ +... + 5⁵⁰) + 5⁵¹ - 5⁰ + (5¹ + 5² + 5³ +... + 5⁴⁹ + 5⁵⁰) = 5⁵¹ - 1.
Tức, A = (5⁵¹ - 1)/4.
Ta có : A = 1 + 5 + 52 + ...... + 549 + 550
=> 5A = 5 + 52 + 53+..... + 550 + 551
=> 5A - A = 551 - 1
=> 4A = 551 - 1
=> \(A=\frac{5^{51}-1}{4}\)
\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\)
5A=\(5+5^2+5^3+...+5^{50}+5^{51}.\)
5A-A=\(\left(5+5^2+5^3+.....+5^{50}+5^{51}\right)-\left(1+5+5^2+5^3+...+5^{49}+5^{50}.\right)\)
4A=\(5^{51}-1\)
\(=>A=\frac{5^{51}-1}{4}\)
\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\)
\(\Rightarrow5A=5+5^2+5^3+...+5^{51}\)
\(\Rightarrow4A=5^{51}-1\)
\(\Rightarrow A=\frac{5^{51}-1}{4}\)
\(5A=5+5^2+5^3+...+5^{50}+5^{51}\)
\(-\)
\(A=1+5+5^2+5^3+...+5^{50}\)
\(4A=5^{51}-1\)
\(A=\frac{5^{51}-1}{4}\)
Đặt A = 1 + 5 + 52 + ....+550
=> 5A = 5 + 52 + ....+551
=> 5A - A = 551 - 1
=> 4A =551 - 1
=> A \(=\frac{5^{51}-1}{4}\)
Đặt A= 1+5+52+...+550
5A=5+52+53+...+551
4A=551-1
A=\(\frac{5^{51}-1}{4}\)
\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\)
=> \(5\text{A}=5+5^2+5^3+5^4...+5^{49}+5^{50}+5^{51}\)
=> \(5\text{A-A}=5+5^2+5^3+5^4...+5^{49}+5^{50}+5^{51}\) - (\(1+5+5^2+5^3+...+5^{49}+5^{50}\) )
=> \(5\text{A-A}=5+5^2+5^3+5^4...+5^{49}+5^{50}+5^{51}\) - \(1-5-5^2-5^3-...-5^{49}-5^{50}\)
=> \(4\text{A}=5^{51}-1\)
=> \(A=\dfrac{5^{51}-1}{4}\)
A=1+\((5+5^2+5^3+...+5^{50})\) 5A=\(5+5^2+5^3+...+5^{51}\) 5A=\((1+5+5^2+...+5^{^{ }50})+5^{51}-1\) 5A=A+\(5^{51}-1\) 5A-A=\((5^{51}-1)\) -A A=\(\dfrac{5^{51-1}}{4}\)
Bài 2:
a: \(=7^4\left(7^2+7-1\right)=7^4\cdot55⋮55\)
b: \(5A=5+5^2+...+5^{51}\)
\(\Leftrightarrow4A=5^{51}-1\)
hay \(A=\dfrac{5^{51}-1}{4}\)
Bài 3:
\(S=\left(1^2+2^3+3^3+...+10^2\right)\cdot2=385\cdot2=770\)
\(5A=5^1+5^2+5^3+...+5^{51}\)
\(4A=5A-A=5^{51}-1\)
\(\Rightarrow A=\frac{5^{51}-1}{4}\)