Question

1 , 3 , 6 , 10 , 15

Answer 1

First difference,

$\Delta_{1} = a_{2} - a_{1} = 3 - 1 = 2 = 1 + 1$

Second difference,

$\Delta_{2} = a_{3} - a_{2} = 6 - 3 = 3 = 2 + 1$

Third difference,

$\Delta_{3} = a_{4} - a_{3} = 10 - 6 = 4 = 3 + 1$

And so on.

Assuming the pattern holds on, we see that

i-th difference,

$\Delta_{i} = a_{i + 1} - a_{i} = i + 1$

$\implies a_{i + 1} = a_{i} + i + 1$

Then, nth term is,

$\implies a_{n} = a_{n - 1} + n$

$= a_{n - 2}+ (n + (n - 1))$

$= a_{n - 3} + (n + (n - 1) +(n - 2))$

$= a_{n - (n - 1)} + \sum \limits^{n - 2}_{k = 0}(n - k)$

$= a_1 + \sum \limits^{n - 2}_{k = 0}n- \sum \limits^{n - 2}_{k = 0}k$

$= a_1 +n(n -2 + 1 )- \frac{1}{2} (n - 2)(n - 1)$

$= a_1 +n(n -1 )- \frac{1}{2} (n - 2)(n - 1)$

$= a_1 +(n -1 )(n- \frac{1}{2} (n - 2))$

$= a_1 + \frac{1}{2} (n -1 )(2n- n + 2)$

$= a_1 + \frac{1}{2} (n -1 )(n + 2)$

$\implies a_{n} = 1 + \frac{1}{2} (n -1 )(n + 2)$

Now, the 21st term in the sequence is,

$\implies a_{21} = 1 + \frac{1}{2} (21 -1 )(21 + 2)$

$= 1 + \frac{1}{2} \times 20 \times 23$

$= 231$