Maclaurin series for $x e^{x^2-1}$

Question

i have done everything to that step$$xe^{x^2-1}=f(x)=\frac 1 e \sum_{n=0}^\infty \frac{x^{2n+1}}{n!},$$ but I need $f^{(101)}(0)$ so I know that I need to find $2n+1=101$ which is $n=50$. But I don't know what to do next. That answer is correct? $$f^{(101)}(0)=\frac 1 e \cdot \frac 1 {50!}$$

Answer 1

Where you wrote $\displaystyle f(n)=\frac 1 e \sum_{i=0}^\infty \frac{x^{2n+1}}{n!}$ you need instead $\displaystyle f(x)=\frac 1 e \sum_{n=0}^\infty \frac{x^{2n+1}}{n!}$ (with $f(x),$ not $f(n),$ and with $\sum\limits_{n=0}^\infty,$ rather than $\sum\limits_{i=0}^\infty).$

The term of degree $101$ is $$ f^{(101)}(0) \frac{x^{101}} {101!} \quad = \quad \frac 1 e \cdot \frac{x^{101}}{50!}, $$ so consequently $$ f^{(101)} (0) \quad = \quad \frac 1 e \cdot \frac{101!}{50!} \quad = \quad \frac{101\cdot100\cdot99\cdot98\cdot97\cdots 51} e. $$

Answer 2

To seek $f^{(101)}(0)$ you need this term from the Maclaurin series: $\frac{f^{(101)}(0)}{101!}x^{101}$.

As you have found, your series has a term with the appropriate power of $x$: $\frac{1}{e}\frac{x^{101}}{50!}$.

So set these things equal and solve for $f^{(101)}(0)$.

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(Also, as @MichaelHardy points out in the comments, a few things are not right with your presentation of the question.)