Question

Find the sequence of differences for the given sequence. Next, find a closed form solution for the following sequence. (Be sure to refer to the example on the modules notes page for “closed form solutions.”)

1, 5, 11, 19, 29, 41, …

Answer 1

Let
`a_n` denote the given sequence with
`n\ge1`:

`\{a_n\}_(n\ge1)=\{1,5,11,19,29,41,\ldots\}`

Let
`b_n` be the sequence of the forward differences of
`a_n`, so that
`b_n=a_(n+1)-a_n` for
`n\ge1`:

`\{b_n\}_(n\ge1)=\{4,6,8,10,12,\ldots\}`

`b_n` follows an arithmetic progression with a difference of 2 between terms, so that

`b_n=4+2(n-1)=2n+2`

Then we have

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

so that
`a_n` is given recursively by

`\begin{cases}a_1=1\\a_(n+1)=a_n+2(n+1)&\text{for }n\ge1\end{cases}`

By substitution, we can try to find a pattern:

`a_2=a_1+2\cdot2`

`a_3=a_2+2\cdot3=a_1+2(2+3)`

`a_4=a_3+2\cdot4=a_1+2(2+3+4)`

`a_5=a_4+2\cdot5=a_1+2(2+3+4+5)`

and so on, with the general pattern

`a_n=a_1+2(2+3+4+\cdots+n)`

and since
`a_1=1` we can write this as

`a_n=1+2(2+3+4+\cdots+n)`

`a_n=-1+2+2(2+3+4+\cdots+n)`

`a_n=2(1+2+3+4+\cdots+n)-1`

Recall that

`\displaystyle\sum_(i=1)^ni=1+2+3+\cdots+n=\frac{n(n+1)}2`

Then

`a_n=2\frac{n(n+1)}2-1\implies\boxed{a_n=n^2+n-1}`