How to prove $\frac{\prod_{k=1}^{31}{}\left(1+\frac{29}{k}\right)}{ \prod_{k=1}^{29}\left(1+\frac{31}{k}\right)}=1$

82 Views Asked by At

Here is a fraction that seems to stump me as to how to work it out. Using a calculator I find out that the answer is 1, and the answer sheet proves me correct. However, if I want to solve future problems like this, I want to know how to solve a question like this without a calculator.

$$\frac{\left ( 1+29\right )\left ( 1+\frac{29}{2}\right )\left (1+ \frac{29}{3} \right )\cdots\left (1+ \frac{29}{30} \right )\left ( 1+\frac{29}{31} \right )}{{\left ( 1+31\right )\left ( 1+\frac{31}{2}\right )\left (1+ \frac{31}{3} \right )\cdots\left (1+ \frac{31}{28} \right )\left ( 1+\frac{31}{29} \right )}}$$

Can someone help me to figure out the workings, please? I would prefer using the simplest method possible, because this is supposed to be a test question and I’m only allowed to spend 5 minutes on this.

1

There are 1 best solutions below

0
On BEST ANSWER

\begin{align} \frac{\prod_{i=1}^{31}(1+\frac{29}{i})}{\prod_{i=1}^{29}1+\frac{31}{i}} &=\left( 1+\frac{29}{30}\right)\left( 1+\frac{29}{31}\right)\prod_{i=1}^{29} \frac{i+29}{i+31}\\ &=\frac{59\cdot 60}{30 \cdot 31}\frac{30\cdot 31}{59 \cdot 60}\\ &=1 \end{align}

Where notice that the numerator is the product $30 \cdot 31 \cdot \ldots \cdot 58$ while the denominator is the product of $32 \cdot \ldots \cdot 59 \cdot 60$.