Study the patterns below and answer the questions that follow. (i) Write down the sixth row of the pattern. (ii) Write down the value of u. (iii) Express w in terms of t.

Here's how to solve the pattern problem:

First, let's find the general formula for each column based on the row number, $n$.

• Column 1: The values are 2, 3, 4, 5, 6 for rows 1, 2, 3, 4, 5 respectively. The pattern is $C_1(n) = n + 1$.

• Column 2: The values are 2, 7, 14, 23, 34. Differences: $7-2=5$, $14-7=7$, $23-14=9$, $34-23=11$. Second differences: $7-5=2$, $9-7=2$, $11-9=2$. Since the second differences are constant (2), the pattern is a quadratic sequence of the form $An^2 + Bn + C$. $2A = 2 \implies A = 1$. $3A + B = 5$ (first difference for $n=1$) $\implies 3(1) + B = 5 \implies B = 2$. $A + B + C = 2$ (value for $n=1$) $\implies 1 + 2 + C = 2 \implies C = -1$. The pattern is $C_2(n) = n^2 + 2n - 1$.

• Column 3: The values are 4, 10, 18, 28, 40. Differences: $10-4=6$, $18-10=8$, $28-18=10$, $40-28=12$. Second differences: $8-6=2$, $10-8=2$, $12-10=2$. Since the second differences are constant (2), the pattern is a quadratic sequence of the form $An^2 + Bn + C$. $2A = 2 \implies A = 1$. $3A + B = 6$ (first difference for $n=1$) $\implies 3(1) + B = 6 \implies B = 3$. $A + B + C = 4$ (value for $n=1$) $\implies 1 + 3 + C = 4 \implies C = 0$. The pattern is $C_3(n) = n^2 + 3n$.