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

1 $\frac{11}{20}$ $\frac{{11}}{{20}}$
2 $\frac{13}{15}$ $\frac{{13}}{{15}}$
3 $\frac{13}{18}$ $\frac{{13}}{{18}}$
4 $\frac{15}{16}$ $\frac{{15}}{{16}}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given by \\ (1 - \frac{1}{3^{2}}) (1 - \frac{1}{4^{2}}) (1 - \frac{1}{5^{2}}) . . . . (1 - \frac{1}{11^{2}}) (1 - \frac{1}{12^{2}}) \\ = [\frac{(3 + 1) (3 - 1)}{(2 + 1) (4 - 1)}] [\frac{(4 + 1) (4 - 1)}{(3 + 1) (5 - 1)}] [\frac{(5 + 1) (5 - 1)}{(4 + 1) (6 - 1)}] . . . . . . . . \\ [\frac{(11 + 1) (11 - 1)}{(10 + 1) (12 - 1)}] [\frac{(12 + 1) (12 - 1)}{(11 + 1) (13 - 1)}] \\ = \frac{(3 - 1)}{(2 + 1)} \times \frac{(12 + 1)}{(13 - 1)} \\ = \frac{2}{3} \times \frac{13}{12} \\ = \frac{13}{18} \end{aligned}

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