Think outside the box (3)

Calculus Level 4

Let $f^{(0)}(x) = (5x^3+99x^{98}+2)^{100}$ , where $f^{(n)}(x)$ denotes the $n^\text{th}$ derivative of $f(x)$ . If the value of $f^{(98)}(0)$ is in the form $(a^2)!\times b^{99}$ , where $a$ and $b$ are natural numbers . Find the value of $a+b$ .

Notation: $!$ is the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$ .

Check out the easier problem .

The answer is 12.

1 solution

Aditya Narayan Sharma
Nov 12, 2016

The multinomial coefficient gives the hint that the general term of the expansion of $(5x^3+99x^{98}+2)^{100}$ is of the form $5^a 99^{b}2^c x^{3a+98b}$ with $3a+98b+c=100$

But we must have $3a+98b=98$ so the only solution is $a=0,b=1,c=99$

So the coefficient of $x^{98}$ is $\displaystyle \frac{100!}{0! 1! 99!}5^0 (99)^1 2^{99}$

Since $\displaystyle \frac{1}{98!}f^{(98)}(0)=100\times 99\times 2^{99} \implies f^{(98)}(0) = (10^2)!\times 2^{99}$

Hence answer is $\boxed{10+2=12}$

Think outside the box (3)

Check out the easier problem .

The answer is 12.

1 solution

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