A<1/1*2+1/2*3+...+1/2021*2022

=>A<1-1/2+1/2-1/3+...+1/2021-1/2022<1

a) \(A=2^1+2^2+2^3+2^4+...+2^{2010}\)

\(A=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)

\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)

\(A=3\left(2+2^3+...+2^{2009}\right)⋮3\)

\(A=2^1+2^2+2^3+2^4+...+2^{2010}\)

\(A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)

\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)

\(A=7\left(2^1+2^4+...+2^{2008}\right)⋮7\)

Các ý dưới bạn làm tương tự nhé. 

cho mik hỏi câu này nữa   a= 2+2 mũ 3 + 2 mũ 5 +.....+2 mũ 51

*Ta có: A\(=2^1+2^2+2^3+2^4+...+2^{2010}\)

              \(=\left(2+2^2\right)+2^2\times\left(2+2^2\right)+...+2^{2008}\times\left(2+2^2\right)\)

              \(=\left(2+2^2\right)\times\left(1+2^2+2^3+...+2^{2008}\right)\)

              \(=6\times\left(2^2+2^3+...+2^{2008}\right)\)

              \(=3\times2\times\left(2^2+2^3+...+2^{2008}\right)\)

               \(\Rightarrow A⋮3\)

*Ta có: A \(=2^1+2^2+2^3+2^4+...+2^{2010}\)

               \(=2\times\left(1+2+2^2\right)+2^4\times\left(1+2+2^2\right)+...+2^{2008}\times\left(1+2+2^2\right)\)

               \(=\left(1+2+2^2\right)\times\left(2+2^4+2^7+...+2^{2008}\right)\)

               \(=7\times\left(2+2^4+2^7+...+2^{2008}\right)\)

                \(\Rightarrow A⋮7\)

Mình sửa lại đề C 1 chút xíu

*Ta có: C \(=3^1+3^2+3^3+3^4+...+3^{2010}\)

               \(=\left(3+3^2\right)+3^2\times\left(3+3^2\right)+...+3^{2008}\times\left(3+3^2\right)\)

               \(=\left(3+3^2\right)\times\left(1+3^2+3^3+...+3^{2008}\right)\)

               \(=12\times\left(1+3^2+3^3+...+3^{2008}\right)\)

               \(=4\times3\times\left(1+3^2+3^3+...+3^{2008}\right)\)

                \(\Rightarrow C⋮4\)

Các câu khác làm tương tự nhé. Chúc bạn học tốt!

Thanks bạn

Giải: 

A= 2 + 2 mũ 2 + 2 mũ 3 + 2 mũ 4 +....+ 2 mũ 2010

A= (2 + 2 mũ 2) + (2 mũ 3 + 2 mũ 4) +....+ (2 mũ 2009 + 2 mũ 2010)

A= 2(1 + 3) + 2 mũ 3 (1 + 2) + 2 mũ 2009 (1 +2_

A= 2.3 + 2 mũ 3.3 +....+ 2 mũ 2009.3

A= 3.(2 + 2 mũ 3 +....+ 2 mũ 2009) chia hết cho 3

A= (2 + 2 mũ 2 + 2 mũ 3) + (2 mũ 4 + 2 mũ 5 + 2 mũ 6) +....+ (2 mũ 2008 + 2 mũ 2009 + 2 mũ 2010)

A= 2(1 + 2 + 2 mũ 2) + 2 mũ 4(1+ 2 + 2 mũ 2) +...+ 2 mũ 2008.(1 + 2 + 2 mũ 2)

A= 2.7 + 2 mũ 4. 7 +.... + 2 mũ 2008.7

A= 7.(2 + 2 mũ 4 +....+ 2 mũ 22010 chia hết cho 7.

Các câu còn lại làm tương tự như câu a nha bạn!

a) \(S=1+5+5^2+5^3+...+5^{28}\)

\(S=\left(1+5\right)+\left(5^2+5^3\right)+...+\left(5^{27}+5^{28}\right)\)

\(S=1\left(1+5\right)+5^2\left(1+5\right)+...+5^{27}\left(1+5\right)\)

\(S=\left(1+5^2+...+5^{27}\right).6⋮3\left(dpcm\right)\)

b) \(S=1+5+5^2+5^3+...+5^{28}\)

\(\Rightarrow5S=5+5^2+5^3+5^4+...+5^{29}\)

\(\Rightarrow5S-S=\left(5+5^2+5^3+5^4+...+5^{29}\right)-\left(1+5+5^2+5^3+...+5^{28}\right)\)

\(\Rightarrow4S=5^{29}-1\)

\(\Rightarrow4S+1=5^{29}-1+1\)

\(\Rightarrow4S=5^{29}=5^n\)

\(\Rightarrow n=29\)

a) \(S=1+5+5^2+5^3+...+5^{28}\)

\(\Rightarrow S=\left(1+5\right)+5^2\left(1+5\right)+...+5^{27}\left(1+5\right)\)

\(\Rightarrow S=6+5^2.6+...+5^{27}.6\)

\(\Rightarrow S=6\left(1+5^2+...+5^{27}\right)⋮6\)

\(\Rightarrow S=6\left(1+5^2+...+5^{27}\right)⋮3\)

\(\Rightarrow dpcm\)

b) Bạn xem lại đề

A=\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+\(\frac{1}{5^2}\)+...+\(\frac{1}{98^2}\)

A=\(\frac{1}{3.3}\)+\(\frac{1}{4.4}\)+\(\frac{1}{5.5}\)+...+\(\frac{1}{98.98}\)

A<\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{97.98}\)=\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+...+\(\frac{1}{97}\)-\(\frac{1}{98}\)=\(\frac{1}{2}\)-\(\frac{1}{98}\)=\(\frac{24}{49}\)<1.

Vậy A<1