Exponents in Fractions

Use $1-\dfrac{1}{x^2}=(1-\dfrac{1}{x})(1+\dfrac{1}{x})$ .

Let $\mathfrak{P}=\displaystyle\prod_{x=2}^m\left(1-\dfrac{1}{x^2}\right)$ $\mathfrak{P}=\left(\dfrac 12\cdot\color{#0C6AC7}{\dfrac 32}\right)\left(\color{#0C6AC7}{\dfrac 23}\cdot\color{#456461}{\dfrac 43}\right)\left(\color{#456461}{\dfrac 34}\cdot\color{goldenrod}{\dfrac 54}\right)\cdots \left(\color{#D61F06}{\dfrac{m-1}{m}}\cdot\dfrac{m+1}{m}\right)$ $\huge =\dfrac{m+1}{2m}$

When $m=2014$ , $\large \mathfrak{P}=\dfrac{2015}{4028}$ $\Large 2015+4028=6043$ $\large \therefore 6+4+0+3=\Huge\color{#456461}{\boxed{\color{#D61F06}{\boxed{\color{#007fff}{13}}}}}$

The given expression can be written as:

$\Huge\prod _{ i=2 }^{ 2014 }{ 1-\frac { 1 }{ { i }^{ 2 } } } =\prod _{ i=2 }^{ 2014 }{ \frac { { i }^{ 2 }-1 }{ { i }^{ 2 } } } =\prod _{ i=2 }^{ 2014 }{ \frac { (i-1)(i+1) }{ i\times i } }$

$\Huge\Rightarrow \frac { \prod _{ i=3 }^{ 2015 }{ i } \prod _{ i=1 }^{ 2013 }{ i } }{ \prod _{ i=2 }^{ 2014 }{ i } \prod _{ i=2 }^{ 2014 }{ i } } =\frac { 2015\times 1 }{ 2\times 2014 } =\frac { 2015 }{ 4028 }$

Hence $x+y=2015+4028=6043$ . Sum of digits of $6043=\color{#3D99F6}{\boxed { 13 }}$