Given: 0; -1; 1; 6; 14; .. - NSC Mathematics - Question 2 - 2016 - Paper 1

From the first differences, we see a pattern. The last first difference is 8, so the next first difference will be ( 8 + 3 = 11 ).

Thus, the next term can be calculated as follows: [ 14 + 11 = 25 ]

Therefore, the next term of the sequence is 25.

Given that the second difference is constant, we can express the nth term as: [ T_n = a n^2 + b n + c ]

We can calculate coefficients a, b, and c using the known terms:

1. For n = 1: ( T_1 = 0 ) => ( a + b + c = 0 )
2. For n = 2: ( T_2 = -1 ) => ( 4a + 2b + c = -1 )
3. For n = 3: ( T_3 = 1 ) => ( 9a + 3b + c = 1 )

Solving these equations, we find: [ a = \frac{3}{2}, b = -\frac{11}{2}, c = 4 ]

Thus, the nth term is: [ T_n = \frac{3}{2}n^2 - \frac{11}{2}n + 4 ]

Using the expression for the nth term: [ T_{30} = \frac{3}{2}(30^2) - \frac{11}{2}(30) + 4 ] Calculating further: [ T_{30} = \frac{3}{2}(900) - \frac{11}{2}(30) + 4 ] [ = 1350 - 165 + 4 ] [ = 1189 ]

Thus, the 30th term is 1189.

From the series: [ a + 13 + b + 27 + ... ] Using the information given:

1. The sum of first two terms is: [ a + 13 = 6 + 13 \Rightarrow a = 6 ]
2. For b, we see: [ b + 27 = 20 + 27 \Rightarrow b = 20 ]

Thus, it is proven that a = 6 and b = 20.

To find the term equal to 230: [ S_n = a + (n-1)d ] where ( a = 6, d = b + 13 - 6 = 20 - 6 = 14 ) Set this equal to 230: [ 6 + (n-1) * 14 = 230 ] [ (n-1) * 14 = 224 ] [ n-1 = 16 \Rightarrow n = 17 ]

Thus, the 17th term of the series is 230.

The series converges if the terms approach zero as ( n \to \infty ).

We can use the convergence test: [ 1 - \frac{k}{5}, ; \left(1 - \frac{k^2}{5}\right), ; \left(1 - \frac{k^3}{5}\right) ] will converge if: [ |1 - \frac{k}{5}| < 1 \Rightarrow |k| < 10 ]

Thus, the series converges for ( k < 10 ).

Using the formula for the sum of the first n terms: [ S_n = \frac{n}{2} (a + l) ] where ( a = 16 ) and ( l = a + (n-1)d ) for the arithmetic series. Calculating:

1. Find d: [ d = 3 - 16 = -13 ]
2. Find l for n = 40: [ l = 16 + (40-1)(-13) = 16 - 507 = -491 ]
3. Thus the sum S_40 is: [ S_{40} = \frac{40}{2} (16 - 491) = 20 * (-475) = -9500 ]

Therefore, the sum of the first 40 terms is -9500.

The series can be written as: [ T_k = 16 \cdot (\frac{1}{2})^{k-1} ] Thus: [ S = \sum_{k=1}^{\infty} T_k = \sum_{k=1}^{\infty} 16 \cdot (\frac{1}{2})^{k-1} ]

To determine ( S_0 ): This series converges to: [ S_0 = \frac{a}{1 - r} ; ext{where} , a = 16, r = \frac{1}{2} ] Therefore: [ S_0 = \frac{16}{1 - \frac{1}{2}} = \frac{16}{\frac{1}{2}} = 32 ]

Hence, ( S_0 = 32 ).