Question

Determine formula of the nth term 2, 6, 12 20 30,42​

Answer 1

Check the forward differences of the sequence.

If
`\{a_n\} = \{2,6,12,20,30,42,\ldots\}`, then let
`\{b_n\}` be the sequence of first-order differences of
`\{a_n\}`. That is, for n ≥ 1,

`b_n = a_(n+1) - a_n`

so that
`\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}`.

Let
`\{c_n\}` be the sequence of differences of
`\{b_n\}`,

`c_n = b_(n+1) - b_n`

and we see that this is a constant sequence,
`\{c_n\} = \{2, 2, 2, 2, \ldots\}`. In other words,
`\{b_n\}` is an arithmetic sequence with common difference between terms of 2. That is,

`2 = b_(n+1) - b_n \implies b_(n+1) = b_n + 2`

and we can solve for
`b_n` in terms of
`b_1=4`:

`b_(n+1) = b_n + 2`

`b_(n+1) = (b_(n-1)+2) + 2 = b_(n-1) + 2*2`

`b_(n+1) = (b_(n-2)+2) + 2*2 = b_(n-2) + 3*2`

and so on down to

`b_(n+1) = b_1 + 2n \implies b_(n+1) = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)`

We solve for
`a_n` in the same way.

`2(n+1) = a_(n+1) - a_n \implies a_(n+1) = a_n + 2(n + 1)`

Then

`a_(n+1) = (a_(n-1) + 2n) + 2(n+1) \\ ~~~~~~~= a_(n-1) + 2 ((n+1) + n)`

`a_(n+1) = (a_(n-2) + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_(n-2) + 2 ((n+1) + n + (n-1))`

`a_(n+1) = (a_(n-3) + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_(n-3) + 2 ((n+1) + n + (n-1) + (n-2))`

and so on down to

`a_(n+1) = a_1 + 2 \displaystyle \sum_(k=2)^(n+1) k = 2 + 2 * \frac{n(n+3)}2`

`\implies a_(n+1) = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}`