What Is the Maximum Value of P(x) in the Sequence?

[anemone]

Let $P(x)=\dfrac{99^x+19^x}{x!}$ for $x=1,\,2,\,3,\cdots$.

Find $x$ such that $P(x)$ is greatest.

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[anemone]

Congratulations to kaliprasad for his correct solution::)

Model Answer:

Spoiler

Notice that $P(k)<P(k+1)$ iff $\dfrac{99^x+19^x}{x!}<\dfrac{99^{x+1}+19^{x+1}}{(x+1)!}$.

Simplifying it gives

$19^x(x-18)+99^x(x-98)<0$, which is true when $x=1,\,2,\,\cdots,97$ and is false for $x\ge 98$.

Hence, $P(k)$ is increasing for $x=1,\,2,\,\cdots,98$ and starts to decrease as $x$ inreases.

Thus, $P(x)$ is greatest at $x=98$