2026 DSS P1 Q16

The first four terms of a sequence are given below.

\(\begin{aligned} T_1&=(2-1)^2+5=6 \\ T_2&=(4-1)^2+9=18 \\ T_3&=(6-1)^2+13=38 \\ T_4&=(8-1)^2+17=66 \end{aligned}\)

1. Find \(T_5\).[1]
2. Show that the \(n\)th term of the sequence, \(T_n\), is given by \(4n^2+2\).[3]
3. \(T_k\) and \(T_{k+1}\) are consecutive terms in the sequence.
   Find and simplify an expression, in terms of \(k\), for \(T_{k+1}-T_k\).[2]
4. Explain why two consecutive terms of the sequence cannot have a difference of 10.[1]