\(1+5+14+30+\ldots \mathrm{n}\) terms \(=\) (A) \([\\{(n+2)(n+3)\\} /(12)]\) (B) \([\\{n(n+1)(n+5)\\} /(12)]\) (C) \([\\{n(n+2)(n+3)\\} /(12)]\) (D) \(\left[\left\\{n(n+1)^{2}(n+2)\right\\} /(12)\right]\)

In this step, we need to identify the pattern that governs the terms of the series. We are given the terms 1, 5, 14, 30, and so on. Let's find the differences between consecutive terms: 5 - 1 = 4 14 - 5 = 9 30 - 14 = 16 The differences between the consecutive terms seem to be perfect squares (1, 4, 9, 16,...). We can see that these are the squares of consecutive positive integers (1, 2, 3,...).

Now that we have identified the pattern in the series, let's analyze the given options and see which one matches the pattern we found. We will substitute the first few values of n into each option and see which one generates the given terms. (A): For n=1, \(\frac{(1+2)(1+3)}{12} = \frac{3 \cdot 4}{12} = 1\) For n=2, \(\frac{(2+2)(2+3)}{12} = \frac{4 \cdot 5}{12} = \frac{10}{3}\) (doesn't match the given terms) (B): For n=1, \(\frac{n(n+1)(n+5)}{12} = \frac{1(1+1)(1+5)}{12} = \frac{1 \cdot 2 \cdot 6}{12} = 1\) For n=2, \(\frac{n(n+1)(n+5)}{12} = \frac{2(2+1)(2+5)}{12} = \frac{2 \cdot 3 \cdot 7}{12} = 3.5\) (doesn't match the given terms) (C): For n=1, \(\frac{n(n+2)(n+3)}{12} = \frac{1(1+2)(1+3)}{12} = \frac{1 \cdot 3 \cdot 4}{12} = 1\) For n=2, \(\frac{n(n+2)(n+3)}{12} = \frac{2(2+2)(2+3)}{12} = \frac{2 \cdot 4 \cdot 5}{12} = 10/3\) (doesn't match the given terms) (D): For n=1, \(\frac{n(n+1)^{2}(n+2)}{12} = \frac{1(1+1)^{2}(1+2)}{12} = \frac{1 \cdot 2^{2} \cdot 3}{12} = 1\) For n=2, \(\frac{n(n+1)^{2}(n+2)}{12} = \frac{2(2+1)^{2}(2+2)}{12} = \frac{2 \cdot 3^{2} \cdot 4}{12} = 5\) For n=3, \(\frac{n(n+1)^{2}(n+2)}{12} = \frac{3(3+1)^{2}(3+2)}{12} = \frac{3 \cdot 4^{2} \cdot 5}{12} = 14\) For n=4,\(\frac{n(n+1)^{2}(n+2)}{12} = \frac{4(4+1)^{2}(4+2)}{12} = \frac{4 \cdot 5^{2} \cdot 6}{12} = 30\) All the terms from option (D) match the given terms in the series so the correct answer is: (D) \(\frac{n(n+1)^{2}(n+2)}{12}\)