There are 64 deer in a forest in the present year. If x(n) denotes the number of deer in the forest in year n and that \(x(n)=\dfrac{3}{2}\,x(n-1)\) for n = 1, 2, 3, ....., how many deer are there in 5th year in the forest ?
486
The relation \(x(n) = \tfrac{3}{2}\,x(n-1)\) means the deer population is multiplied by \(\tfrac{3}{2}\) every year, which is a geometric progression. With \(x(0) = 64\) for the present year, the general term is \(x(n) = 64 \left(\tfrac{3}{2}\right)^{n}\).
Step 1 — \(x(1) = 64 \times \tfrac{3}{2} = 96\).
Step 2 — \(x(2) = 96 \times \tfrac{3}{2} = 144\), \(x(3) = 144 \times \tfrac{3}{2} = 216\).
Step 3 — \(x(4) = 216 \times \tfrac{3}{2} = 324\), \(x(5) = 324 \times \tfrac{3}{2} = 486\).
Step 4 — Check with the closed form: \(x(5) = 64 \times \frac{243}{32} = 2 \times 243 = 486\).
The value 96 is the population after only one year. The value 243 is \(3^5\), the numerator of \(\left(\tfrac{3}{2}\right)^{5}\) without the factor \(\tfrac{64}{32} = 2\). The value 339 does not arise from repeated multiplication by \(\tfrac{3}{2}\) at any stage.
486 option is correct.
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