1 + 2 + 4 + 8 + 16 + 32 + ... + 1024 + 2048 + 4096 =?
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1+1=2
2+2=4
4+4=8
8+8=16
16+16=32
64+64=128
128+128=256
512+512=1024
2048+2048=4096
xong
(2+4+6+8+...+98+100)x 10/2+4+8+16+...+512-2
Ta có:
+) 2 + 4 + 6 + 8 + ... + 98 + 100
= (100 + 2)[(100 - 2) : 2 + 1] : 2
= 102.50:2 = 2550
+)Dặt A = 2 + 4 + 8 + 16 + ... + 512 - 2
A = 2 + 2^2 + 2^3 + 2^4 + ... + 2^9 - 2
A = 2^2 + 2^3 + 2^4 + ... + 2^9
2A = 2(2^2 + 2^3 + 2^4 + .... + 2^9)
2A = 2^3 + 2^4 + 2^5 + ... + 2^10
2A - A = (2^3 + 2^4 + 2^5 + ... + 2^10) - (2^2 + 2^3 + 2^4 + ... + 2^9)
A = 2^10 - 2^2
A = 1024 - 4 = 1020
=> (2 + 4 + 6 + 8 + ... + 98 + 100) x 10/2 + 4 + 8 + 16 + ... + 512 - 2
= 2550 x 10/1020 = 25500/1020 = 25
Đặt
\(S=1+2+4+...+2048+4096\)
\(S=1+2^1+2^2+...+2^{11}+2^{12}\)
\(2S=2+2^2+2^3+...+2^{12}+2^{13}\)
\(2S-S=\left(2+2^2+2^3+...+2^{13}\right)-\left(1+2+2^2+..+2^{12}\right)\)
\(S=2^{13}-1=8192-1=8191\)
Gọi A=1+2+4+8+16+...+1024+2048+4096
2A=2+4+8+16+32+...+2048+4096+8192
2A-A=(2+4+8+16+32+...+2048+4096+8192)-(1+2+4+8+16+...+1024+2048+4096)
A=8192-1
A=8191
Đặt A=1/2+1/4+1/8+1/16+1/32+...+1/2048+1/4096
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{12}}\)
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{12}}\right)\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{11}}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{11}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{12}}\right)\)
\(A=1-\frac{1}{2^{12}}\)
Đặt A=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+................+\frac{1}{2048}+\frac{1}{4096}\)
2A=\(1+\frac{1}{2}+\frac{1}{4}+....................+\frac{1}{1024}+\frac{1}{2048}\)
2A-A=\(\left(1+\frac{1}{2}+\frac{1}{4}+................+\frac{1}{1024}+\frac{1}{2048}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...............+\frac{1}{2048}+\frac{1}{4096}\right)\)
A=\(1-\frac{1}{4096}\)
A=\(\frac{4095}{4096}\)
Từ biểu thức trên ta được
A=1/2+1/2^2+1/2^3+....+1/2^11+1/2^12
2A-A=2(1/2+1/2^2+1/2^3+....+1/2^11+1/2^12)-(1/2+1/2^2+1/2^3+....+1/2^11+1/2^12)
A=1+1/2^2+1/2^3+....+1/2^11-1/2-1/2^2-...-1/2^12
A=1/2-1/2^12
Ủng hộ cho mình nha bạn
\(A=\)\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}+\frac{1}{2048}+\frac{1}{4096}\)
\(A=\)\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{12}}\)
\(2A=\)\(2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{12}}\right)\)
\(2A=\)\(1+\frac{1}{2}+...+\frac{1}{2^{11}}\)
\(2A-A=\)\(\left(1+\frac{1}{2}+...+\frac{1}{2^{11}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{12}}\right)\)
\(A=1-\frac{1}{2^{12}}\)
1+1=2
2+2=4
4+4=8
8+8=16
16+16=32
32+32=64
64+64=128
128+128 = 256
256+256=512
512+512= 1024
1024+1024 = 2048
2048 + 2048 = 4096
1+1=2
2+2=4
4+4=8
8+8=16
16+16=32
32+32=64
64+64=128
128+128=256
256+256=512
512+512=1024
1024+1024=2048
2048+2048=4096
Đặt \(A=1+2+4+.........+4096\)
\(2A=2+4+8+......+8192\)
\(\Rightarrow2A-A=8192-1\)
\(\Rightarrow A=8191\)
Đặt \(S=1+2+4+...+1024+2048+4096\)
\(S=1+2^1+2^2+2^3+....+2^{10}+2^{11}+2^{12}\)
\(2S=2+2^2+2^3+....+2^{11}+2^{12}+2^{13}\)
\(2S-S=\left(2+2^2+2^3+....+2^{12}+2^{13}\right)-\left(1+2+2^2+....+2^{11}+2^{12}\right)\)
\(S=2^{13}-1=8192-1=8191\)