So big, yet so small

Algebra Level 2

$\large \left(1+\dfrac{1}{\color{#D61F06}1}\right)\left(1+\dfrac{1}{\color{#D61F06}2} \right)\left(1+\dfrac{1}{\color{#D61F06}3}\right)\left(1+\dfrac{1}{\color{#D61F06}4}\right)\ldots \left(1+\dfrac{1}{\color{#D61F06}{2016}}\right) = \ ?$

The answer is 2017.

4 solutions

Rohit Udaiwal
Nov 29, 2015

$\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2} \right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)\ldots \left(1+\dfrac{1}{2016}\right) \\ =\dfrac{2}{1} \cdot \dfrac{3}{2} \cdot \dfrac{4}{3} \cdot \dfrac{5}{4} \ldots \cdot \dfrac{2017}{2016} \\ =2017.$

Following a similar method, can you generalize the solution for $\displaystyle\mathcal{P}(m,n):=\prod_{k=m}^n\left(1+\frac 1k\right)$ where $m,n\in\Bbb Z^+$ and $m\leq n$ ?

Then, the current problem would just be evaluating $\mathcal{P}(1,2016)$ .

Prasun Biswas - 5 years, 6 months ago

Sadasiva Panicker
Dec 9, 2015

2/1 x 3/2 x 4/3 x.........x3017/2016 = 2017

Aly O'Mahoney
Dec 19, 2015

2/1 × 3/2 × 4/3 ... × 2017/2016 Every number can cancel out (fractionally) apart from the first '1' and the last '2017' which leaves 2017/1 = 2017

Mohammed Marzuq Rahman
Nov 18, 2020

Observe $1 + \frac {1}{n}$ $=$ $\frac{n+1}{n}$ $--(1)$
So lets approach the problem using the idea above.

It can be seen all the numerator and denominator are being cancelled except denominator of 1st fraction= 1 and numerator of last fraction = 2017
So answer is $\frac{2017}{1}=\boxed{2017}$

So big, yet so small

The answer is 2017.

4 solutions

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