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cho P=1/2+1/3+1/4+...........+1/48+1/49+1/50 và Q=1/49+2/48+3/47+........+47/3+48/2+49/1
tao có : P= 1/49+2/48+3/47+...+48/2+{1+1+...+1}(49 số 1)
P= (1/49+1)+(2/48+1)+(3/47+1)+...+(48/2+1)+1
P= 50/49+50/48+50/47+...+50/2+50/50
P=50.(1/2+1/3+1/4+...+1/48+1/49+1/50)}S
=> P/S=50/1
Q = \(\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+\frac{49}{1}\)
Cộng 1 vào mỗi phân số trong 48 phân số đầu, trừ phân số cuối đi 48, ta được :
Q = \(\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(\frac{3}{47}+1\right)+...+\left(\frac{48}{2}+1\right)+1\)
Q = \(\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+...+\frac{50}{2}+1\)
Q = \(\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+...+\frac{50}{2}+\frac{50}{50}\)
đưa phân số cuối lên đầu :
Q = \(\frac{50}{50}+\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+...+\frac{50}{2}\)
Q = \(50.\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+...+\frac{1}{2}\right)\)
Q = 50 . A
Vậy \(\frac{P}{Q}=\frac{1}{50}\)
So sánh tổng : S = 1/5 + 1/9 + 1/10 + 1/41 + 1/42 với 1/2
S=
=50/50+50/49+50/48+...+50/2
=50.(1/50+1/49+1/48+...+1/4+1/3+1/2)
=50
P=
P=(1/49+1)+(2/48+1)+...+(48/2+1)+1
P= 50/49+50/48+....+50/2+50/50=1
vậy s/p = 1/50
mức trung bình của năm số tự nhiên khác nhau là 4 nếu sự khác nhau giữa lớn nhất và nhỏ hơn của những con số này là càng lớn càng tốt với mức trung bình của số cây khác là những gì
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\\ 2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\\ 2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\\ A=1-\dfrac{1}{2^{100}}\)
\(E=\dfrac{3^2}{2\cdot4}+\dfrac{3^2}{4\cdot6}+...+\dfrac{3^2}{198\cdot200}\\ =3^2\cdot\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{198\cdot200}\right)\\ =9\cdot\dfrac{1}{2}\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{198\cdot200}\right)\\ =\dfrac{9}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{198}-\dfrac{1}{200}\right)\\ =\dfrac{9}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{200}\right)\\ =\dfrac{9}{2}\cdot\dfrac{99}{200}\\ =\dfrac{891}{400}\)
\(B=1\cdot2+2\cdot3+3\cdot4+...+48\cdot49\\ 3B=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot4\cdot3+...+48\cdot49\cdot3\\ 3B=1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+48\cdot49\cdot\left(50-47\right)\\ 3B=1\cdot2\cdot3-0\cdot1\cdot2+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+48\cdot49\cdot50-47\cdot48\cdot49\\ 3B=-0\cdot1\cdot2+\left(1\cdot2\cdot3-1\cdot2\cdot3\right)+\left(2\cdot3\cdot4-2\cdot3\cdot4\right)+...+\left(47\cdot48\cdot49-47\cdot48\cdot49\right)+48\cdot49\cdot50\\ =0+48\cdot49\cdot50\\ =48\cdot49\cdot50\\ B=\dfrac{48\cdot49\cdot50}{3}\\ B=39200\)
C1:
\(C=1^2+2^2+3^2+...+48^2\\ =1^2+1+2^2+2+3^2+3+...+48^2+48-1-2-3-...-48\\ =1\cdot\left(1+1\right)+2\cdot\left(2+1\right)+3\cdot\left(3+1\right)+...+48\cdot\left(48+1\right)-\left(1+2+3+...+48\right)\\ =1\cdot2+2\cdot3+3\cdot4+...+48\cdot49-\left(1+2+3+...+48\right)\)
Gọi \(S=1\cdot2+2\cdot3+3\cdot4+...+48\cdot49\)
\(S=1\cdot2+2\cdot3+3\cdot4+...+48\cdot49\\ 3S=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot4\cdot3+...+48\cdot49\cdot3\\ 3S=1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+48\cdot49\cdot\left(50-47\right)\\ 3S=1\cdot2\cdot3-0\cdot1\cdot2+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+48\cdot49\cdot50-47\cdot48\cdot49\\ 3S=-0\cdot1\cdot2+\left(1\cdot2\cdot3-1\cdot2\cdot3\right)+\left(2\cdot3\cdot4-2\cdot3\cdot4\right)+\left(3\cdot4\cdot5-3\cdot4\cdot5\right)+...+\left(47\cdot48\cdot49-47\cdot48\cdot49\right)+48\cdot49\cdot50\\ 3S=0+48\cdot49\cdot50\\ 3S=48\cdot49\cdot50\\ S=\dfrac{48\cdot49\cdot50}{3}\\ S=39200\)
\(C=S-\left(1+2+3+...+48\right)\\ C=39200-\left(\dfrac{48\cdot49}{2}\right)\\ C=39200-1176\\ C=38024\)