101+102+103.+199+200=? 101+102+103...+199+200=?

101+102+103.+199+200=? 101+102+103...+199+200=?

101+102+103...+199+200=100*100+(1+2+...+99+100)=10000+5050=15050

100/101+102/101+103/102+.199/200=?

100/101+102/101+103/102+.199/200
=1- 1/101 + 1 + 1/101 +.1-1/200
=100-1/200
=99.995

101-102+103-104.199-200=?

Group summation. It's two arithmetic sequences. The difference is 2. Put the positive ones together and the negative ones together. Gauss addition. You will. I think it's better than telling you the answer

(101 + 103 +... + 199) / (90 + 92 +... + 188) simple operation

The sum of arithmetic sequence Sn = (a1 + an) * n / 2 has the same number of numerator and denominator terms n, so the original formula = (101 + 199) / (90 + 188) = 1.079

If the quotient of B divided by a is 0.35, then the simplest integer ratio of a and B is ()

A: B = 1:0.35 = 100:35 = 20:7

The quotient of a divided by B is 0.625, and the simplest integer ratio of a and B is 0.625

Number A: number b = 0.625:1 = 625:1000 = 5:8

The quotient of a divided by B is 0.25. What is the simplest integer ratio of a and B

Number a △ number b = 0.25 = 25 / 100 = 1 / 4 = 1:4

1 / 2 + 5 / 6 + 11 / 12 + 19 / 20 + +9890 out of 9900

Is the title correct, 9899
=(1-1/2)+(1-1/6)+(1-1/12))+(1-1/20))+(1-))+(1-1/42)、、、+)+(1-1/9900)
Take the denominator as 1 × 2, 2 × 3, 3 × 4, 99 × 100, there are 99 formulas
=99-（1/2+1/6+1/12+1/20+1/30、、、+1/9900）
=99-[1/(1×2)+1/(2×3)+1/(3×4)+、、、+1/(99×100)]
=99-(1-1/2+1/2-1/3+1/3-1/4+、、、+1/99-1/100)
=99-(1-1/100)
=99-99/100
=98.01

12+56+1112+1920+2930+… +97019702+98999900．

12+56+1112+1920+2930+… +97019702+98999900=（1-12）+（1-16）+（1-112）+… +（1-19900）=1×99-（12+16+112+… +19900），=99-（11×2+12×3+13×4… +199×100），=99-（1-12+12-13+13−14+… +199-1100），=99-（1-1100），=99-99100，=981100．

1 and 1 / 3 + 7 / 12-9 / 20 + 11 / 30-13 / 42 = how much?

=4/3+(1/3+1/4)-(1/4+1/5)+(1/5+1/6)-(1/6+1/7)
=4/3+1/3+1/4-1/4-1/5+1/5+1/6-1/6-1/7
=5/3-1/7
=32/21
=One and 21, 11