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

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

Then

$a_n=2\dfrac{n(n+1)}2-1\implies\boxed{a_n=n^2+n-1}$