JAMB 1994 · UME · Q21

Find the $n$ th term of the sequence $3, 6, 10, 15, 21, \dots$

Worked solution (try it first)

Each term is half the product of two consecutive numbers: $3 = \frac{2 \times 3}{2}$ , $6 = \frac{3 \times 4}{2}$ , $10 = \frac{4 \times 5}{2}$ .
The first factor is always one more than the term number, so the $n$ th term is $\frac{(n + 1)(n + 2)}{2}$ .
Check $n = 5$ : $\frac{6 \times 7}{2} = 21$ .
So the answer is option C.