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\(P=\left(1^2+2^2+...............+2015^2\right):\left(2^2+4^2+........+4030^2\right)\)
\(P=\left(1^2+2^2+............+2015^2\right):\left[\left(1.2\right)^2+\left(2.2\right)^2+.............+\left(2.2015\right)^2\right]\)
\(P=\left(1^2+2^2+........+2015^2\right):\left(1^2.2^2+2^2.2^2+...............+2015^2.2^2\right)\)
\(P=\left(1^2+2^2+......+2015^2\right):2^2.\left(1^2+2^2+.........+2015^2\right)\)
\(P=\left(1^2+2^2+........+2015^2\right).\frac{1}{2^2.\left(1^2+2^2+..............+2015^2\right)}\)
\(P=\frac{1^2+2^2+...............+2015^2}{2^2.\left(1^2+2^2+............+2015^2\right)}=\frac{1}{2^2}=\frac{1}{4}\)
Chúc bạn học tốt
Ta có A = 1 + 2 + 22 + 23 + ... + 2100
=> 2A = 2 + 22 + 23 + 24 + ... + 2101
Khi đó 2A - A = (2 + 22 + 23 + 24 + ... + 2101) - (1 + 2 + 22 + 23 + ... + 2100)
=> A = 2101 - 1
Vì 2101 - 1 < 2101
=> A < B
Vậy A < B
A = 1 + 2 + 22 + 23 + ... + 2100
=> 2A = 2( 1 + 2 + 22 + 23 + ... + 2100 )
= 2 + 22 + 23 + ... + 2101
=> A = 2A - A
= 2 + 22 + 23 + ... + 2101 - ( 1 + 2 + 22 + 23 + ... + 2100 )
= 2 + 22 + 23 + ... + 2101 - 1 - 2 - 22 - 23 - ... - 2100
= 2101 - 1 < 2101
=> A < B
#)Giải : (Đg rảnh nên làm lun :v)
Ta có : \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}=1-\frac{1}{51}=\frac{50}{51}< 2\)
\(\Rightarrow A< \frac{50}{51}< 2\)
\(\Rightarrow A< 2\left(đpcm\right)\)
\(D=1+3+3^2+3^3+...+3^{2002}\)
\(3D=3.\left(1+3+3^2+3^3+...+3^{2002}\right)\)
\(=3+3^2+3^3+3^4+...+3^{2003}\)
\(3D-D=\left(3+3^2+3^3+3^4+...+3^{2003}\right)-\left(1+3+3^2+3^3+...+3^{2002}\right)\)
\(2D=3^{2003}-1\)
\(D=\frac{3^{2003}-1}{2}\)
\(D=1+3+3^2+3^3+..........+3^{2002}\)
\(\Rightarrow3D=3+3^2+3^3+3^4+......+3^{2003}\)
\(\Rightarrow3D-D=2D=3^{2003}-1\)
\(\Rightarrow D=\frac{3^{2003}-1}{2}\)
Dễ thấy \(\frac{1}{2^2}< \frac{1}{1\cdot2};...;\frac{1}{100^2}< \frac{1}{99\cdot100}\)
Do đó : \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
Ta có : \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}221+321+421+...+10021<1.21+2.31+3.41+...+99.1001
=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1=1−21+21−31+31−41+...+991−1001=1−1001<1
\(a,[\left(8.x-12\right):4].3^3.3=3^6.6\)
\(\left(8x-12\right):4=54\)
\(8x-12=216\)
\(8x=228\)
\(x=28,5\)
\(b,41-2^{x+1}=9\)
\(2^{x+1}=41-9\)
\(2^{x+1}=32\)
\(2^{x+1}=2^5\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=4\)
thanks bạn nhìu :)))