por favor ayudenm
Realizar los siguientes ejercicios de sumatorias y productorias. Se debe colocar el desarrollo y resultado del operador. miren aki esta la imagen...
Adjuntos:
![](https://es-static.z-dn.net/files/dc5/891c40f2db8ac5d4fdbcaaa093562b9b.png)
Respuestas
Respuesta dada por:
2
A)
; Desarrollamos la sumatoria:
![= (2(1)+1)^{2}+(2(2)+1)^{2}+(2(3)+1)^{2}+(2(4)+1)^{2}+(2(5)+1)^{2} = (2(1)+1)^{2}+(2(2)+1)^{2}+(2(3)+1)^{2}+(2(4)+1)^{2}+(2(5)+1)^{2}](https://tex.z-dn.net/?f=%3D+%282%281%29%2B1%29%5E%7B2%7D%2B%282%282%29%2B1%29%5E%7B2%7D%2B%282%283%29%2B1%29%5E%7B2%7D%2B%282%284%29%2B1%29%5E%7B2%7D%2B%282%285%29%2B1%29%5E%7B2%7D)
![= (2+1)^{2}+(4+1)^{2}+(6+1)^{2}+(8+1)^{2}+(10+1)^{2} = (2+1)^{2}+(4+1)^{2}+(6+1)^{2}+(8+1)^{2}+(10+1)^{2}](https://tex.z-dn.net/?f=%3D+%282%2B1%29%5E%7B2%7D%2B%284%2B1%29%5E%7B2%7D%2B%286%2B1%29%5E%7B2%7D%2B%288%2B1%29%5E%7B2%7D%2B%2810%2B1%29%5E%7B2%7D)
![= (3)^{2}+(5)^{2}+(7)^{2}+(9)^{2}+(11)^{2} = (3)^{2}+(5)^{2}+(7)^{2}+(9)^{2}+(11)^{2}](https://tex.z-dn.net/?f=%3D+%283%29%5E%7B2%7D%2B%285%29%5E%7B2%7D%2B%287%29%5E%7B2%7D%2B%289%29%5E%7B2%7D%2B%2811%29%5E%7B2%7D)
![= 9+25+49+81+121 = 9+25+49+81+121](https://tex.z-dn.net/?f=%3D+9%2B25%2B49%2B81%2B121)
![=285 =285](https://tex.z-dn.net/?f=%3D285)
Por tanto:![\Sigma ^{5}_{i=1} (2i+1)^{2}=285 \Sigma ^{5}_{i=1} (2i+1)^{2}=285](https://tex.z-dn.net/?f=+%5CSigma+%5E%7B5%7D_%7Bi%3D1%7D+%282i%2B1%29%5E%7B2%7D%3D285)
B)
; Desarrollamos la Multiplicatoria:
![\Pi ^{4}_{i=1} \frac{i}{i+1} = (\frac{1}{1+1})(\frac{2}{2+1})(\frac{3}{3+1})(\frac{4}{4+1}) \Pi ^{4}_{i=1} \frac{i}{i+1} = (\frac{1}{1+1})(\frac{2}{2+1})(\frac{3}{3+1})(\frac{4}{4+1})](https://tex.z-dn.net/?f=+%5CPi+%5E%7B4%7D_%7Bi%3D1%7D++%5Cfrac%7Bi%7D%7Bi%2B1%7D+%3D+%28%5Cfrac%7B1%7D%7B1%2B1%7D%29%28%5Cfrac%7B2%7D%7B2%2B1%7D%29%28%5Cfrac%7B3%7D%7B3%2B1%7D%29%28%5Cfrac%7B4%7D%7B4%2B1%7D%29+)
; Simplificando:
![=(\frac{1}{1})(\frac{1}{1})(\frac{1}{1})(\frac{1}{5}) =(\frac{1}{1})(\frac{1}{1})(\frac{1}{1})(\frac{1}{5})](https://tex.z-dn.net/?f=%3D%28%5Cfrac%7B1%7D%7B1%7D%29%28%5Cfrac%7B1%7D%7B1%7D%29%28%5Cfrac%7B1%7D%7B1%7D%29%28%5Cfrac%7B1%7D%7B5%7D%29)
![= \frac{1}{5} = \frac{1}{5}](https://tex.z-dn.net/?f=%3D+%5Cfrac%7B1%7D%7B5%7D+)
Por tanto:![\Pi ^{4}_{i=1} \frac{i}{i+1}= \frac{1}{5} \Pi ^{4}_{i=1} \frac{i}{i+1}= \frac{1}{5}](https://tex.z-dn.net/?f=%5CPi+%5E%7B4%7D_%7Bi%3D1%7D+%5Cfrac%7Bi%7D%7Bi%2B1%7D%3D+%5Cfrac%7B1%7D%7B5%7D+)
Por tanto:
B)
Por tanto:
vilbox:
oye muchasssss pero muchas gracias, gente como tu es la que necesitamos cada ves mas :D gracias de nuevo
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