The value of $$\left( {1 - \frac{1}{{{3^2}}}} \right)$$ $$\left( {1 - \frac{1}{{{4^2}}}} \right)$$ $$\left( {1 - \frac{1}{{{5^2}}}} \right)$$   . . . . . $$\left( {1 - \frac{1}{{{{11}^2}}}} \right)$$ $$\left( {1 - \frac{1}{{{{12}^2}}}} \right)$$   $$ = ?$$

Solution(By Examveda Team)

  $$\left( {1 - \frac{1}{{{3^2}}}} \right)$$ $$\left( {1 - \frac{1}{{{4^2}}}} \right)$$ $$\left( {1 - \frac{1}{{{5^2}}}} \right)$$   . . . . . $$\left( {1 - \frac{1}{{{{11}^2}}}} \right)$$ $$\left( {1 - \frac{1}{{{{12}^2}}}} \right)$$
  $$ = \left( {\frac{{{3^2} - 1}}{{{3^2}}}} \right)\left( {\frac{{{4^2} - 1}}{{{4^2}}}} \right)\left( {\frac{{{5^2} - 1}}{{{5^2}}}} \right)....$$       $$\left( {\frac{{{{11}^2} - 1}}{{{{11}^2}}}} \right)$$ $$\left( {\frac{{{{12}^2} - 1}}{{{{12}^2}}}} \right)$$
  $$ = \left[ {\frac{{\left( {3 + 1} \right)\left( {3 - 1} \right)}}{{\left( {2 + 1} \right)\left( {4 - 1} \right)}}} \right]$$   $$\left[ {\frac{{\left( {4 + 1} \right)\left( {4 - 1} \right)}}{{\left( {3 + 1} \right)\left( {5 - 1} \right)}}} \right]$$   $$\left[ {\frac{{\left( {5 + 1} \right)\left( {5 - 1} \right)}}{{\left( {4 + 1} \right)\left( {6 - 1} \right)}}} \right]$$   . . . . . $$\left[ {\frac{{\left( {11 + 1} \right)\left( {11 - 1} \right)}}{{\left( {10 + 1} \right)\left( {12 - 1} \right)}}} \right]$$   $$\left[ {\frac{{\left( {12 + 1} \right)\left( {12 - 1} \right)}}{{\left( {11 + 1} \right)\left( {13 - 1} \right)}}} \right]$$
$$\eqalign{ & = \frac{{\left( {3 - 1} \right)}}{{\left( {2 + 1} \right)}} \times \frac{{\left( {12 + 1} \right)}}{{\left( {13 - 1} \right)}} \cr & = \frac{2}{3} \times \frac{{13}}{{12}} \cr & = \frac{{13}}{{18}} \cr} $$