Question

$f\left(x\right)=(x^2+8x+20)e^x$. Find $\left[\frac{f^{100}\left(0\right)}{100f\left(0\right)}\right]$ and $f^n\left(x\right)$ denotes the nth derivative of f.
(The notation [x] denotes the greatest integer that is less than or equal to x.)

Answer 1

We have

$f\left(x\right)=(x^2+8x+20)e^x$

On differentiation $f\left(x\right)$ wrt $x$

$f^1\left(x\right)=(2x+8)e^x+(x^2+8x+20)e^x$

$f^2\left(x\right)=2e^x+2\times (2x+8)e^x+(x^2+8x+20)e^x$

$f^3\left(x\right)=2e^x(1+2)+3\times (2x+8)e^x+(x^2+8x+20)e^x$

$f^4\left(x\right)=2e^x(1+2+3)+4\times (2x+8)e^x+(x^2+8x+20)e^x$

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$f^n\left(x\right)=2e^x(1+2+3+\cdots +(n-1\left)\right)+n\times (2x+8)e^x+(x^2+8x+20)e^x$

$\therefore$$f^{100}\left(x\right)=2e^x(1+2+3+\cdots +(100-1\left)\right)+100\times (2x+8)e^x+(x^2+8x+20)e^x$

$f^{100}\left(x\right)=2e^x(99\times 50)+100(2x+8)e^x+(x^2+8x+20)e^x$

$f^{100}\left(0\right)=2(99\times 50)+100\left(8\right)+20=100\times 107+20=10720$

$\therefore [\frac{f^{100}\left(0\right)}{100f\left(0\right)}]=[\frac{10720}{2000}]=\left[5.36\right]=5$