Tính tổng sau :
D = 1/1x3 + 1/3x5 + .... + 1/(2n-1)x(2n+1)
E = 1/1x3x5 + 1/3x5x7 + .... + 1/(2n-1)x(2n+1)x(2n+3)
mk cần gấp lắm
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1/1x3 + 1/3x5 + 1/5x7 + ... + 1/(2n+1)x(2n+3) = n+1/2n+3
2/1x3 + 2/3x5 + 2/5x7 + ... + 2/(2n+1)x(2n+3) = 2n+2/2n+3
1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2n+1 - 1/2n+3 = 2n+2/2n+3
1 - 1/2n+3 = 2n+2/2n+3
Bn nào thông minh thế, ra bài này đố Tây lm đc, ai lm đc mk bái lm sư phụ lun, sửa đề đê
Ủng hộ mk nha ^_-
a, Đặt :
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+..............+\dfrac{1}{19.21}\)
\(\Leftrightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+............+\dfrac{2}{19.21}\)
\(\Leftrightarrow2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+..........+\dfrac{1}{19}-\dfrac{1}{21}\)
\(\Leftrightarrow2A=1-\dfrac{1}{21}\)
\(\Leftrightarrow2A=\dfrac{20}{21}\)
\(\Leftrightarrow A=\dfrac{10}{21}\)
b, \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...........+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Leftrightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+............+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Leftrightarrow2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\)
\(\Leftrightarrow2A=1-\dfrac{1}{2n+1}\)
\(\Leftrightarrow2A=\dfrac{2n}{2n+1}\)
\(\Leftrightarrow A=\dfrac{n}{2n+1}\)
\(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{\left(2n+1\right)x\left(2x+3\right)}=\frac{n+1}{2n+3}\)
=>\(2x\left(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{\left(2n+1\right)x\left(2n+3\right)}\right)=2x\frac{n+1}{2n+3}\)
=>\(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{\left(2n+1\right)\left(2n+3\right)}=\frac{2n+2}{2n+3}\)
=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n+1}-\frac{1}{2n+3}=\frac{2n+2}{2n+3}\)
=>\(1-\frac{1}{2n+3}=\frac{2n+2}{2n+3}\)
=>\(\frac{2n+2}{2n+3}=\frac{2n+2}{2n+3}\)
=>.....
1)5(x^2-1)+x(1-5x)= x-2
<=>5x2-5+x-5x2=x-2
<=>-5+x=x-2
<=>x-x=-2+5
<=>0x=3(vô lí)
vậy ko tìm được x
Dãy số có 2 chữ số chia hết cho 3 là:[12,15,....,99]
Khoảng cách của từng số hạng là 3
Số số hạng là: (99-12):3+1=30(số)
Vậy có 30 số có 2 chữ số chia hết cho 3
B=1+32+33+...+3100
3B=3(1+32+33+...+3100)
3B=3+33+...+3101
3B-B=(3+33+...+3101)-(1+32+33+...+3100)
2B=3101-1
\(B=\frac{3^{101}-1}{2}\)
1 . tìm giá trị x
\(\left(x+1\right)+\left(x+4\right)+....+\left(x+28\right)=115.\)
\(\Rightarrow\left(x+x+x+....x\right)+\left(1+4+..+28\right)=115\)
\(\Rightarrow10x+\left(28+1\right).10:2=115\)
\(\Rightarrow10x+145=115\)
\(\Rightarrow10x=115-145=-30\)
\(\Rightarrow x=-30:10=-3\)
Bai 1
so so hang la: [(28+x)-(x+1)]/3+1= 10 so hang
tong =[(x+1)+(x+28)]*10/2=(2x+29)*10/2=115
(2x+29)*5=115
2x+29=115/5=23
2x=23-29=-6
x=-3
\(D= \dfrac{1}{1.3} + \dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right).\left(2n+1\right)}\),
\(2.D = \dfrac{2}{1.3}+ \dfrac{2}{3.5}+...+\dfrac{2}{\left(2n-1\right).\left(2n+1\right)}\)
\(2.D = 1 - \dfrac{1}{3} + \dfrac{1}{3}- \dfrac{1}{5} +\dfrac{1}{5}- \dfrac{1}{7} + ... + \dfrac{1}{\left(2n-1\right)}-\dfrac{1}{\left(2n+1\right)}\)
\(2.D = 1 - \dfrac{1}{\left(2n+1\right)}\)
\(2.D= \dfrac{2n}{\left(2n+1\right)} \)
Vậy \(D = \dfrac{n}{\left(2n+1\right)}\)
\(E=\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+\dfrac{1}{5.7.9}+...+\dfrac{1}{\left(2n-1\right).\left(2n+1\right).\left(2n+3\right)}\)
\(\Rightarrow4E=4.\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+\dfrac{1}{5.7.9}+...+\dfrac{1}{\left(2n-1\right).\left(2n+1\right).\left(2n+3\right)}\)
\(=\dfrac{4}{1.3.5}+\dfrac{4}{3.5.7}+...+\dfrac{4}{\left(2n-1\right).\left(2n+1\right).\left(2n+3\right)}\)
\(=\dfrac{1}{1.3}-\dfrac{1}{3.5}+\dfrac{1}{3.5}-\dfrac{1}{5.7}-...+\dfrac{1}{\left(2n-1\right).\left(2n+1\right)}-\dfrac{1}{\left(2n+1\right).\left(2n+3\right)}\)
\(=\dfrac{1}{1.3}-\dfrac{1}{\left(2n+1\right).\left(2n+3\right)}\)
\(\Rightarrow E=\dfrac{\dfrac{1}{1.3}-\dfrac{1}{\left(2n+1\right).\left(2n+3\right)}}{4}\)
\(=\dfrac{1}{12}-\dfrac{1}{\left(2n+1\right).\left(2n+3\right).4}\)