Quotient of Complicated Products

Algebra Level 3

For each positive integer $n$ , let $a_n=\dfrac{n^2}{2n+1}$ . Furthermore, let $P$ and $Q$ be real numbers such that $P=a_1a_2\ldots a_{2013},\qquad Q=(a_1+1)(a_2+1)\ldots(a_{2013}+1).$ What is the sum of the digits of $\frac QP$ ?

The answer is 31.

2 solutions

Cody Johnson
Apr 16, 2014

$\begin{aligned} \frac{\prod_{n=1}^{2013}(a_n+1)}{\prod_{n=1}^{2013}a_n}&=\prod_{n=1}^{2013}\frac{a_n+1}{a_n}\\ &=\prod_{n=1}^{2013}\frac{\frac{n^2}{2n+1}+1}{\frac{n^2}{2n+1}}\\ &=\prod_{n=1}^{2013}\frac{\frac{n^2+2n+1}{2n+1}}{\frac{n^2}{2n+1}}\\ &=\prod_{n=1}^{2013}\frac{n^2+2n+1}{n^2}\\ &=\prod_{n=1}^{2013}\frac{(n+1)^2}{n^2}\\ &=\frac{\prod_{n=1}^{2013}(n+1)^2}{\prod_{n=1}^{2013}n^2}\\ &=\frac{\prod_{n=2}^{2014}n^2}{\prod_{n=1}^{2013}n^2}\\ &=\frac{2014^2\cdot\prod_{n=2}^{2013}n^2}{1^2\cdot\prod_{n=2}^{2013}n^2}\\ &=2014^2=4056196 \end{aligned}$

$4+0+5+6+1+9+6=\boxed{31}$

An easy way to compute $2014^2$ is $2000^2+14^2+2(2000)(14)=4000000+196+56000=4056196$

Cody Johnson - 7 years, 1 month ago

Cool problem! !!!!

Manu Attri - 7 years ago

Finn Hulse
Apr 17, 2014

Exactly the same as @Cody Johnson . Great job! :D

Quotient of Complicated Products

The answer is 31.

2 solutions

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