Tinutulungan ka ng Imhr.ca na makahanap ng maaasahang mga sagot sa lahat ng iyong mga katanungan mula sa mga eksperto. Tuklasin ang libu-libong tanong at sagot mula sa isang komunidad ng mga eksperto sa aming madaling gamitin na platform. Kumonekta sa isang komunidad ng mga eksperto na handang magbigay ng eksaktong solusyon sa iyong mga tanong nang mabilis at eksakto.
Sagot :
Solution:
[tex]\sf \sum\limits_{x=0}^{\infty}\big(\frac{2^x(x^2+x-1)(x^2+x+1)}{e^2x!}\big)[/tex]
[tex]= \sf \sum\limits_{x=0}^{\infty}\big(\frac{e^{-2}2^x(x^2+x-1)(x^2+x+1)}{x!}\big)[/tex]
[tex]= \sf \sum\limits_{x=0}^{\infty}\big(\frac{e^{-2}2^x(x^4+2x^3+x^3-1)}{x!}\big)[/tex]
[tex]= \sf e^{-2} \sum\limits_{x=0}^{\infty}\big(\frac{2^x(x^4+2x^3+x^3-1)}{x!}\big)[/tex]
[tex]= \sf e^{-2} \big[\sum\limits_{x=0}^{\infty}\big(\frac{2^x(x^4)}{x!}\big) + \sum\limits_{x=0}^{\infty}\big(\frac{2^x(2x^3)}{x!}\big) + \sum\limits_{x=0}^{\infty}\big(\frac{2^x(x^2)}{x!}\big) - \sum\limits_{x=0}^{\infty}\big(\frac{2^x(1)}{x!}\big)\big][/tex]
[tex]= \sf (e^{-2})(94e^2 + 44e^2 + 6e^2-e^2)[/tex]
[tex]= \sf (e^{-2})(143e^{2})[/tex]
[tex]= \sf 143e^{0}[/tex]
[tex]= \large \boxed{\sf 143}[/tex]
??
[tex] \large \bold{SOLUTION:} [/tex]
[tex] \!\!\begin{array}{l} \small \textsf{Let } S = \displaystyle \sum_{x=0}^{\infty}\frac{2^x(x^2+x-1)(x^2+x+1)}{e^2x!} \\ \\ \small S =\displaystyle e^{-2} \sum_{x=0}^{\infty}\frac{2^x((x^2+x)^2 -1)}{x!} \\ \\ \small S = \displaystyle e^{-2} \sum_{x=0}^{\infty}\frac{2^x(x^4 +2x^3 + x^2 -1)}{x!} \\ \\ \small{S} = \displaystyle \footnotesize e^{-2}\! \left( \sum_{x=0}^{\infty}\frac{2^x x^4}{x!} + 2 \sum_{x=0}^{\infty}\frac{2^x x^3}{x!} + \sum_{x=0}^{\infty} \frac{2^x x^2}{x!} - \sum_{x=0}^{\infty} \frac{2^x}{x!}\right) \end{array} [/tex]
[tex]\!\! \small \begin{array}{l} \textsf{The above series can be evaluated by the Recurrence} \\ \textsf{Relation} \\ \\ \qquad \large \displaystyle \sum_{x=0}^{\infty} \frac{z^x x^k}{x!} = z \frac{d}{dz} \sum_{x=0}^{\infty} \frac{z^x x^{k-1}}{x!} \\ \\ \textsf{Starting from }k = 0, \\ \\ \bullet \displaystyle \sum_{x=0}^{\infty} \frac{z^x}{x!} = e^z \\ \\ \bullet \displaystyle \sum_{x=0}^{\infty} \frac{z^x x}{x!} = z\frac{d}{dz} e^z = ze^z \\ \\ \bullet \displaystyle \sum_{x=0}^{\infty} \frac{z^x x^2}{x!} = z\frac{d}{dz} ze^z = z(ze^z + e^z) = (z^2 + z)e^z \end{array} [/tex]
[tex]\!\!\small \begin{array}{l} \begin{aligned} \bullet \displaystyle \sum_{x=0}^{\infty}\frac{z^x x^3}{x!} &= z\frac{d}{dz} (z^2 + z)e^z \\ \\ &= z(z^2 + z + 2z + 1)e^z \\ \\ &= (z^3 + 3z^2 + z)e^z \end{aligned} \\ \\ \begin{aligned} \bullet \displaystyle \sum_{x=0}^{\infty}\frac{z^x x^4}{x!} &= z\frac{d}{dz}(z^3 + 3z^2 + z)e^z \\ \\ & = z(z^3 + 3z^2 + z + 3z^2 + 6z + 1)e^z \\ \\ &= (z^4 + 6z^3 + 7z^2 + z)e^z \end{aligned} \end{array} [/tex]
[tex] \small \begin{array}{l} \textsf{Evaluating at }z = 2, \\ \\ \bullet \displaystyle \sum_{x=0}^{\infty} \frac{2^x}{x!} = e^2 \\ \\ \bullet \displaystyle \sum_{x=0}^{\infty} \frac{2^x x^2}{x!} = (2^2 + 2)e^2 = 6e^2 \\ \\ \bullet \displaystyle \sum_{x=0}^{\infty}\frac{2^x x^3}{x!} = (2^3 + 3\cdot 2^2 + 2)e^2 = 22e^2 \\ \\ \bullet \displaystyle \sum_{x=0}^{\infty}\frac{2^x x^4}{x!} = (2^4 + 6\cdot 2^3 + 7\cdot 2^2 + 2)e^2 = 94e^2 \\ \\ \\ \implies S = e^{-2} e^{2} (94 + 2\cdot 22 + 6 - 1) \\ \\ \implies S = \boxed{143} \end{array} [/tex]
[tex] \mathfrak{\#Brainlèss\_Squad} [/tex]
Salamat sa pagpili sa aming serbisyo. Kami ay nakatuon sa pagbibigay ng pinakamahusay na mga sagot para sa lahat ng iyong mga katanungan. Bisitahin muli kami. Pinahahalagahan namin ang iyong oras. Mangyaring bumalik anumang oras para sa pinakabagong impormasyon at mga sagot sa iyong mga tanong. Maraming salamat sa paggamit ng Imhr.ca. Bumalik muli para sa karagdagang kaalaman mula sa aming mga eksperto.
