2010-II-12 – Solving Master

Ans: D
First, rewrite the sequence as follow:

$\frac{0}{3}, \frac{- 1}{4}, \frac{2}{5}, \frac{- 3}{6}, \frac{4}{7}, \dots$

Then, we have

$\begin{array}{rcl} T (1) & = & \frac{0}{3} \\ T (2) & = & \frac{- 1}{4} \\ T (3) & = & \frac{2}{5} \\ T (4) & = & \frac{- 3}{6} \\ T (5) & = & \frac{4}{7} \\ ⋮ \end{array}$

Hence by comparing the term number and the term itself, $T (n) = (- 1)^{n + 1} \frac{n - 1}{n + 2}$ .

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