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1)  \(5^1+5^2+5^3+...+5^{2003}+5^{2004}=\) \(\left(5^1+5^4\right)+\left(5^2+5^5\right)+\left(5^3+5^6\right)+...+\left(5^{2001}+5^{2004}\right)\)

\(=5\left(1+5^3\right)+5^2\left(1+5^3\right)+5^3\left(1+5^3\right)+...+5^{2001}\left(1+5^3\right)\)

\(=\left(1+5^3\right).\left(5+5^2+5^3+...+5^{2001}\right)\)

\(=126.\left(5+5^2+5^3+...+5^{2001}\right)⋮126\) \(\left(đpcm\right)\)

Mình biết đáp án rồi nhưng vẫn viết đáp án cho các bạn khác :
Đáp án : 1200

là 1200 đúng ko 

duyệt nha

Bạn biết đáp án rồi sao vẫn còn đăng ??????????????????????????????????????????????

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{49\cdot50}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-.....+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{1}{2}-\frac{1}{50}\)

\(=\frac{24}{50}=\frac{12}{25}\)

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{49\cdot50}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{1}{2}-\frac{1}{50}\)

\(=\frac{12}{25}\)

      \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{1}{2}-\frac{1}{50}\)

\(=\frac{25}{50}-\frac{1}{50}\)

\(=\frac{24}{50}=\frac{12}{25}\)

A = 1 + 2 + 2² + ... + 2¹⁰⁰

=> 2A = 2 + 2² + ... + 2¹⁰¹

=> 2A - A = 2¹⁰¹ - 1

=> A = 2¹⁰¹ - 1

Theo bài ra ta có :

2⁵⁰.(2¹⁰¹-1+1) = 2^m

=> 2⁵⁰.2¹⁰¹ = 2^m

=> 2¹⁵¹ = 2^m

=> m = 151

cứ mỗi p/số kia bé hơn:1+1/1.2+1/2.3+1/3.4+....+1/49.50

phân phối ra nhé còn:2-1/50

mà 1/50>0

=>A<2

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

A=\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}<\frac{1}{1.1}+\frac{1}{1.2}+....+\frac{1}{49.50}\)

A=\(\frac{1}{1}-\frac{1}{50}=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}<2=\frac{2}{1}\)

A=\(\frac{49}{50}<\frac{2}{1}=\frac{49}{50}<\frac{100}{50}\)

Vậy A<2 hay\(\frac{49}{50}<2\)

A=\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}\)

A=\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}<\frac{1}{0.1}+\frac{1}{1.2}+...+\frac{1}{49.50}\)

A=\(\frac{1}{1}-\frac{1}{50}=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}\)

A=\(\frac{49}{50}<2=\frac{49}{50}<\frac{100}{50}\)

Vậy A<2 hay \(\frac{49}{50}<\frac{2}{1}\)