Descendant Series

Calculus Level 3

$\large Z(x) = \sum_{n=1}^\infty n(n-1)(n-2) x^{n-3}$

Let $Z(x)$ be a function as described above for $|x| < 1$ . If the prime factorization of $Z\left(\dfrac12\right)$ is $a^c \times b$ , where $a,b$ and $c$ are prime numbers , find $\dfrac{a+b}c$ .

Give your answer to 1 decimal place.

Hint : The following convergent geometric series may prove useful: $\displaystyle \sum_{n=1}^\infty x^n= \dfrac1{1-x}$ .

The answer is 1.0.

1 solution

David Hontz
May 21, 2016

Notice that $\frac{d^3}{dx^3} x^n = n(n-1)(n-2)x^{n-3}$ ; therefore, $Z(x) = \sum_{n=1}^{\infty} n(n-1)(n-2)x^{n-3} =\frac{d^3}{dx^3}\sum_{n=1}^{\infty} x^n =\frac{d^3}{dx^3} (\frac{1}{1-x}) = \frac{6}{(1-x)^4}$ $Z(\frac{1}{2}) =\frac{6}{(1-\frac{1}{2})^4}=\frac{6}{(\frac{1}{2})^4} = 6(2^4) = 6(16) = 96 =2^5\times 3$ $\boxed{a=2; b=3; c=5}$ $Thus: \frac{a+b}{c} = \frac{2+3}{5} = \frac{5}{5} = \boxed{1.0}$

Nice problem! The $n(n-1)(n-2)$ strongly suggests the use of differentiation.

展豪張 - 5 years ago

Thank you for the kind words. I'm glad you saw the hint for differentiation in the formula since it was never verbally stated in the problem itself. Excellent observation!

David Hontz - 5 years ago

Awesome problem!

Jakob Evanoski - 4 years, 12 months ago

Descendant Series

The answer is 1.0.

1 solution

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