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]