Question

Question #b65ad

Answer 1

The `n^"th"` term is given by `(n-1)(n+1)`

Explanation:

Examining the sequence, we find that the difference between successive terms increase by `2`, starting with an increase of `3`. That is, we have

`a_0 = 0`
`a_1 = 0 + 3`
`a_2 = 0 + 3 + 5`
`a_3 = 0 + 3 + 5 + 7`
`...`

As the difference increases by `2` each time, we can see that for each `k>=1`, the difference between `a_(k-1)` and `a_k` is `3 + 2(k-1)`. Thus, noting that the `n^"th"` term is `a_(n-1)` because we started from `a_0`, we have

`a_(n-1) = 0 + 3 + 5 + ... + (3+2((n-1)-1))`

`=(3+2(0))+(3+2(1))+...+(3+2(n-2))`

`=(3+3+...+3) + (2(0)+2(1) + ... + 2(n-2))`

`=3(n-1) + 2(1+2+...+(n-2))`

`=3(n-1)+2(((n-2)(n-1))/2)`

(The above comes from that the sum of all successive integers from `1` to `k` is given by `(k(k+1))/2`, with `k=n-2` in this case)

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

`=(n-1)(n+1)`

Checking we find that this matches up with the given values. E.g. the `3^"rd"` term is `8 = 2*4 = (3-1)(3+1)`