The value of ( 1 - 13^2 ) ( 1 - 14^2 ) ( 1 - 15^2 ) . . . . . ( 1 - 111^2 ) ( 1 - 112^2 ) = ?

Answer: Option C

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