Question

Find the $n^{\text{th}}$ term of the sequence 2, 6, 12, 20 ...

• A. $n^2$ - n
• B. n$\times$(n+1)
• C. (n+1)$\times$(n-1)
• D. n$\times$(n-2)

Answer 1

The correct option is B n$\times$(n+1)

$T_2$ = $T_1$ + 4 = 2 + 4 = 2$\times$3 = $2^2$ + 2 = 6
$T_3$ = $T_2$ + 6 = 6 + 6 = 3$\times$4 = $3^2$ + 3 = 12
$T_4$ = $T_3$ + 8 = 10 + 8 = 4$\times$5 = $4^2$ + 4 = 20


similarly $T_n$ = n$\times$(n+1) or $n^2$ + n