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Đặt \(A=2^{100}-2^{99}+2^{98}-2^{97}+...-2^3+2^2\)
\(=\left(2^{100}-2^{99}\right)+\left(2^{98}-2^{97}\right)+...+\left(2^4-2^3\right)+4\)
\(=2^{99}\left(2-1\right)+2^{97}\left(2-1\right)+...+2^3\left(2-1\right)+4\)
\(=\left(2^3+2^5+...+2^{97}+2^{99}\right)+4\)
Đặt \(B=2^3+2^5+...+2^{97}+2^{99}\)
\(\Rightarrow4B=2^5+2^7+...+2^{99}+2^{101}\)
\(\Rightarrow4B-B=3B=2^{101}-2^3\)
\(B=\frac{2^{101}-2^3}{3}\)
\(A=B+4\)
\(=\frac{2^{101}-2^3}{3}+4\)
\(=\frac{2^{101}-8+12}{3}\)
\(=\frac{2^{101}+4}{3}\)
1 cách khác
M = 299 + 2 . 298 + 3 . 297 + 4 . 296 + ... + 98 . 22 + 99 . 2 + 100 . 20
M = 299 + 2 . ( 299 - 298 ) + 3 . ( 298 - 297 ) + 4 . ( 297 - 296 ) + ... + 99 . ( 22 - 2 ) + 100 . ( 2 - 1 )
M = 299 + 2100 - 2 . 298 + 3 . 298 - 3 . 297 + 4 . 297 - 4. 296 + ... + 99 . 22 - 99 . 2 + 100 . 2 - 100
M = 2100 + 299 +298 + 297 + 296 + ... + 2 - 100
M = 2101 - 102
J=6 + 16 + 30 + 48 +...+ 19600 + 19998
Chia cả 2 vế cho 2 ta được
B/2 = 3 + 8 + 15 + 24 + ......... + 98000+ 9999
B/2= 1x3+2x4+3x5+4x6+…….+98x100+99x101
B/2= 100/6[(100-1)x(2x100+1)] = 328350
-> B =328350x2=656700
K=2 + 5 + 9 + 14 + ....+ 4949 + 5049
Nhân cả 2 vế với 2 ta được
2xD=1x4+ 2x5+ 3x6+ 4x7+……..+98x101+99x102
2xD = 1(2+2)+2(3+2)+3(4+2)+...+99(100+2)
2xD = 1x2+1x2+2x3+2x2+3x4+3x2+...+99x100+99x2
2xD= (1x2+2x3+3x4+...+99x100)+2(1+2+3+...+99)
2xD = 333300 + 9900 = 343200
-> D= 343200 :2 =171600
\(A=2^{100}-2^{99}-...-2-1=2^{100}-\left(2^{99}+2^{98}+...+2+1\right)\)
Đặt \(B=2^{99}+2^{98}+...+2+1\)
\(2B=2^{100}+2^{99}+...+2^2+2\)
\(\Rightarrow2B-B=2^{100}-1\)
\(\Rightarrow B=2^{100}-1\)
\(\Rightarrow A=2^{100}-B=2^{100}-\left(2^{100}-1\right)=1\)
Đặt \(A=2^{100}-2^{99}-2^{98}-...-2^2-2-1\)
=>\(A=2^{100}-\left(2^{99}+2^{98}+...+2^2+2+1\right)\)
Đặt \(B=2^{99}+2^{98}+...+2^2+2+1\)
=>\(2B=2^{100}+2^{99}+...+2^3+2^2+2\)
=>\(2B-B=2^{100}+2^{99}+...+2^3+2^2+2-2^{99}-2^{98}-...-2^2-2-1\)
=>\(B=2^{100}-1\)
\(A=2^{100}-B=2^{100}-2^{100}+1=1\)