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\(A=1+2+2^2+2^3+...+2^{100}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{200}+2^{201}\)
\(2A-A=A=\left(2+2^2+2^3+...+2^{201}\right)-\left(1+2+2^2+2^3+...+2^{200}\right)\)
\(\Rightarrow A=2^{201}-1\Rightarrow A+1=2^{201}-1+1=2^{201}\)
ta có
A= 1+2+22+23+...+2200
2A= 2+22+23+...+2201
2A-A=(2+22+23+...+2201)-(1+2+22+23+...+2200)
A=2201 - 1
A+1=2201
A Chia hết cho 5
A = 20 + 21 + 22 + ....+ 299
=> ( 20 +22 ) +.....+ ( 297 + 299 )
=> 1 ( 1 + 4 ) + ... + 297 ( 1 + 4 )
1 . 5 +.....+ 297 . 5
5 ( 1 + .... + 297 ) chia hết cho 5
A=22+22+23+24+.........+22005
\(2A=2^3+2^3+2^4+2^5+...+2^{2006}\)
\(2A-A=\left(2^3+2^3+2^4+2^5+...+2^{2006}\right)-\left(2^2+2^2+2^3+2^4+...+2^{2005}\right)\)
\(A=\left(2^{2006}+2^3\right)-\left(2^2+2^2\right)\)
\(A=\left(2^{2006}+2^3\right)-2^3\)
\(A=2^{2006}\)
\(A=2^2+2^2+2^3+2^4+...+2^{2005}\)
\(\Rightarrow2A=2^3+2^3+2^4+2^5+...+2^{2005}+2^{2006}\)
\(\Rightarrow2A-A=\left(2^3+2^3+2^4+2^5+...+2^{2006}\right)-\left(2^2+2^2+2^3+2^4+...+2^{2005}\right)\)
Triệt tiêu hai vế \(\Rightarrow A=\left(2^{2006}+2^3\right)-\left(2^2+2^2\right)=2^{2006}+2^3-2^3\)
\(\Rightarrow A=2^{2006}\)
1) A = 1+2+222 + ... + 22002200
2A = 2 + 222 + 233 + ... + 2201201
2A - A = 2 + 222 +233 + ... + 22012201 - 1 - 2 - ... - 2200200
A = 2201201 - 1
A+1 = 2201201
Vậy a + 1 = 2201201
2) C = 3 + 322 + 333 + ... + 320052005
3C = 322 + 333 + 344 + ... + 320062006
3C - C = 3232 + 333 + 344 + ... + 320062006 - 3 - 322 - 333 - ... - 320052005
2C = 320062006 - 3
2C+3 = 320062006
Vậy 2C + 3 là luỹ thừa của 3 ( Đpcm )
Bài 1:
Ta có: \(A=1+2+2^2+...+2^{200}\)
\(\Leftrightarrow2A=2+2^2+2^3+...+2^{201}\)
\(\Rightarrow2A-A=\left(2+2^2+...+2^{201}\right)-\left(1+2+...+2^{200}\right)\)
\(\Leftrightarrow A=2^{201}-1\)
\(\Rightarrow A+1=2^{201}\)
Bài 2:
Ta có: \(B=3+3^2+3^3+...+3^{2005}\)
\(\Leftrightarrow3B=3^2+3^3+3^4+...+3^{2006}\)
\(\Rightarrow3B-B=\left(3^2+3^3+...+3^{2006}\right)-\left(3+3^2+...+3^{2005}\right)\)
\(\Leftrightarrow2B=3^{2006}-3\)
\(\Rightarrow2B+3=3^{2006}\)