Question

A playground is being designed where children can interact with their friends in certain combinations. . If there is 1 child, there can be 0 interactions. . If there are 2 children, there can be 3 interactions. . If there are 3 children, there can be 9 interactions. . If there are 4 children, there can be 18 interactions. . Which recursive equation represents the pattern? (2 points)

Answer 1

The correct answer is:

The general term would be given by: $T_n=\frac{3}{2}(n-2)(n-1)$

Explanation:

We first find the first differences between the terms of this sequence:

3-0 = 3; 9-3 = 6; 18-9 = 9

Since these are not the same, we must find the second differences (differences between the first diferences):

6-3 = 3; 9-6 = 3

these are the same, so this will be a quadratic sequence. This will be of the form $T_n=an^2+bn+c$. The value of a will be half of the second difference, which is 3; this means a will be 3/2.

Using the term numbers as n, for the first term, we have:

$T_1=\frac{3}{2}(1^2)+b(1)+c=0 \\ \\=\frac{3}{2}+b+c=0$

For term 2,

$T_2=\frac{3}{2}(2^2)+b(2)+c=0 \\ \\=\frac{3}{2}(4)+2b+c=0 \\ \\=\frac{12}{2}+2b+c=0 \\ \\=6+2b+c=0$

Taking these as a system of equations gives us:

$\left \{ {{\frac{3}{2}+b+c=0} \atop {6+2b+c=0}} \right. \\ \\\text{Subtracting the second equation from the first gives us:} \\ \\ \left \{ {{\frac{3}{2}+b+c=0} \atop {-(6+2b+c=0)}} \right. \\ \\ \text{This gives us:} \\ \\-4.5-b=0 \\ \\\text{Solving this for b, we add 4.5 to each side:} \\ \\-4.5-b+4.5=0+4.5 \\-b=4.5 \\ \\\text{Divide-b both sides by -1:} \\\frac{-b}{-1}=\frac{4.5}{-1} \\b=-4.5$

This gives us:

$T_n=\frac{3}{2}n^2-\frac{9}{2}n+c=0 \\ \\\text{For term #1, we have} \\ \\T_1=\frac{3}{2}(1^2)-\frac{9}{2}(1)+c=0 \\T_1=\frac{3}{2}-\frac{9}{2}+c=0 \\\frac{-6}{2}+c=0 \\-3+c=0 \\-3+c+3=0+3 \\c=3$

This gives us:

$T_n=\frac{3}{2}n^2-\frac{9}{2}n+3$

This factors to:

$T_n=\frac{3}{2}(n-2)(n-1)$

Answer 2

We are given with:when n=1, a(1) = 0 n=2, a(2) =3 n=3, a(2) =9 n=4, a(2) =18The equation that suits the conditions is a(n) = a(n-1) + (n-1)*3