1*3*1/5+3*5*1/7+···+ 2001*2003*1/2005

1*3*1/5+3*5*1/7+···+ 2001*2003*1/2005

1 +3 +5 +7 + .2005
Solution: the original formula = [(1 + 2005) △ 2]
=1003 square meters

1+2-3-4+5+6-7-8+.+2001+2002-2003-2004+2005+2006

=1+(2-3-4+5)+(6-7-8+9)+...+(2002-2003-2004+2005)+2006
=1+0+0+...+0+2006
=2007

1 / 9 / 2 [32.4-3 / 1 × 25%)] × 12

Original = 10 / 9 × (32.4-32.4 + 1 / 3 × 1 / 4) × 12
＝10/9÷(0＋1/12)×12
＝10/9÷1/12×12
＝10/9×12×12
＝160

[(-1)^3÷2/9+2^2001×(-0.5)^2001-3^3×(-3/2)^2]÷│-4÷2×(-1/2)^2│

Original formula = - 1 △ 2 / 9 + 2x (- 0.5) ^ 2001-9x4 / 9 △ 1 / 8
=9/2+(-1)-162
=158.5

9, 3, 1, 3, 1, 3, 8, 2, 8, 32, 9, 4, 15, 6, 18 () 1, half, 2, 1, 3, 3 (), () Six of seven, two of seven, two of twenty-one, () two hundred and eighty-nine () (a title is a line) Seven tenths of the road has been built () × 7 / 10 = ()

9, 3, 1, 3, 1 / 3, 8, 2, 2, 8, 2, 2, 6, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 1, 2, 2, 2, 2, 2, 2, 2, 1, 3, 3, 1, 3, 1, 2, 3, 3, 3, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2

Calculate the ratio of 1,32 / 9:82,24.3,3.5

32 / 9: 8 / 3 = 32 / 9x3 / 8 = 4 / 3
I didn't write all the back ones
If there is anything you don't understand, you can ask,

1/5×12÷1/5= 0.25×12.5÷1/32= 2/3×4/9× 1/5×12÷1/5= 0.25×12.5÷1/32= 2/3×4/9×1/2= 6.9+3.1-6.9+3.1= 1÷1/2-1/2÷1= 3/5+3/5÷3/5+3/5= 1÷1/7×1÷1/7= 4/5×4/5÷4/5×4/5=

Twelve
One hundred
7/27
Six point two
One point five
11/5
Forty-nine
16/25

1 + 3 = 4,1 + 3 + 5 = 9,1 + 3 + 5 + 7 = 16,1 + 3 + 5 + 7 +. + 1997 + 1999 + 2001 + 2003 The sum of 1 + 3 + 5 +. + (2n-5) + (2n-3) + (2n-1) + (2n + 1) is? (where n is a natural number) 1 + 3 = 4 is the square of 2, 1 + 3 + 5 is the square of 3

1+3+5+7+.+1997+1999+2001+2003=[(2003+1)/2]^2=1002^2
1+3+5+.+(2n-5)+(2n-3)+(2n-1)+(2n+1)=(n+1)^2

1+3+5+7+… +1999+2001．

1+3+5+7+… +1999+2001，
=（1+1999）+（3+1997）+… +2001，
=2000×500+2001，
=1002001．

1.3.5.7.9.11.1993.1995.1997.1999.2001.2003.2005. 1.4.7.10.13.1996.1999.2002.2005, how many numbers appear in the two groups at the same time?

The total number of eligible (2005-1) 6 + 1 = 335 can be contacted by message