A4. What is the value of $\frac{\left(1^{2}+1\right)\left(2^{2}+1\right)\left(3^{2}+1\right)}{\left(2^{2}-1\right)\left(3^{2}-1\right)\left(4^{2}-1\right)}$ ? Give your answer in its simplest form.
SolUTiON $\square$
The value of $\frac{\left(1^{2}+1\right)\left(2^{2}+1\right)\left(3^{2}+1\right)}{\left(2^{2}-1\right)\left(3^{2}-1\right)\left(4^{2}-1\right)}$ is $\frac{2 \times 5 \times 10}{3 \times 8 \times 15}=\frac{5 \times 4 \times 5}{3 \times 2 \times 4 \times 3 \times 5}=\frac{5}{3 \times 2 \times 3}$.
So the required value is $\frac{5}{18}$.