It looks like \(e^x\)

Calculus Level 3

If the value of \(\displaystyle\sum_{n=0}^{\infty}\frac{n^2x^n}{n!}\) is \((ax^2+bx+c)e^x\), where \(a\), \(b\), and \(c\) are constants, enter the value of \(123a+456b+789c\).

Additional formula:

  • \(\displaystyle e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\)
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