We multiply both sides of e^x = \sum_{n = 0}^{\infty} \frac{x^n}{n!} \space \text{by} \space (x^2 + x)ex=∑n=0∞n!xn by (x2+x) to get
\displaystyle (x^2 + x) e^x = \sum_{n = 0}^{\infty} (x^2 + x) \frac{x^n}{n!} . (x2+x)ex=n=0∑∞(x2+x)n!xn.
Manipulating the right hand side of the above series yields
\displaystyle \sum_{n = 0}^{\infty} (x^2 + x) \frac{x^n}{n!} = \sum_{n = 0}^{\infty} \frac{x^{n + 2}}{n!} + \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n!}n=0∑∞(x2+x)n!xn=n=0∑∞n!xn+2+n=0∑∞n!xn+1
\displaystyle = x + \sum_{n = 2}^{\infty} \left( \frac{x^n}{(n - 2) !} + \frac{x^n}{(n - 1) !} \right)=x+n=2∑∞((n−2)!xn+(n−1)!xn)
\displaystyle = x + \sum_{n = 2}^{\infty} \left( \frac{n (n - 1) x^n}{n!} + \frac{n x^n}{n!} \right)=x+n=2∑∞(n!n(n−1)xn+n!nxn)
\displaystyle = x + \sum_{n = 2}^{\infty} \frac{n^2 x^n}{n!} = \sum_{n = 1}^{\infty} \frac{n^2 x^n}{n!} = \sum_{n = 0}^{\infty} \frac{n^2 x^n}{n!} .=x+n=2∑∞n!n2xn=n=1∑∞n!n2xn=n=0∑∞n!n2xn.
해설을 그대로 ctrl+C ctrl+V 한것입니다. 어느정도 해석을 하다가도 막히는 부분이있습니다
감사합니다.
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