Question

CUAL ES LA SUMA DE LOS 20 PRIMEROS NÚMEROS DE M
M=3/2+7/6+13/12+21/20+31/30

Answer 1

Valor de M:

Respuesta:

M= $11\frac{1}{4}$

Explicación paso a paso:

$\frac{3}{2}$ + $\frac{7}{6}$ + $\frac{13}{2}$ + $\frac{21}{20}}$ + $\frac{31}{30}$

$(1)= \frac{3}{2}+\frac{7}{6}= 2\frac{2}{3}$

$\frac{3}{2} = 1\frac{1}{2}$

$\frac{7}{6} = 1\frac{1}{6}$

$1\frac{1}{2} + 1\frac{1}{6} + 1\frac{(1)(3)}{6}+ 1\frac{(1) (1)}{6} = 1 + 1 \frac{3 + 1}{6} = 2\frac{2}{3}$

$2\frac{4}{6} = 2\frac{(2)(2)}{(2)(3)} = 2\frac{2}{3}$

$(2) =2\frac{2}{3} + \frac{13}{2} = 9\frac{1}{6}$  

$\frac{13}{2} = 6\frac{1}{2}$

$2\frac{2}{3} + 6\frac{1}{2} = 2\frac{(2) (2)}{6}+6\frac{(1) (3)}{6} =2+6\frac{4+3}{6} = 8\frac{7}{6} = 9\frac{1}{6}$

$(3) = 9\frac{1}{6} +\frac{21}{20} = 10\frac{13}{60}$

$\frac{21}{20} = 1\frac{1}{20} \\\\6= (2) (3)$

$20= /2) (2) (5)$

$MCD: (20) (3) = 60$

$9\frac{1}{6} +1\frac{1}{20}= 9\frac{(1) (10)}{60}+ 1\frac{(1) (3)}{60}= 9+1\frac{10+3}{60}=10\frac{13}{60}$

$(4)=10\frac{13}{60} +\frac{31}{30} =11\frac{1}{4}$

$\frac{31}{30} =\frac{1}{30}$

$10\frac{30}{60} +1\frac{1}{30}=10\frac{(13)(1)}{60} +1\frac{(1)(2)}{60}= 10+1\frac{13+2}{60} =11\frac{15}{60}=11\frac{1}{4}$

$11=\frac{15}{60}=11\frac{(3)(5)}{(2)(2)(3)(5)} =11\frac{1}{4}$