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, …

Let [tex]a_n[/tex] denote the given sequence with [tex]n\ge1[/tex]:[tex]\{a_n\}_{n\ge1}=\{1,5,11,19,29,41,\ldots\}[/tex]Let [tex]b_n[/tex] be the sequence of the forward differences of [tex]a_n[/tex], so that [tex]b_n=a_{n+1}-a_n[/tex] for [tex]n\ge1[/tex]:[tex]\{b_n\}_{n\ge1}=\{4,6,8,10,12,\ldots\}[/tex][tex]b_n[/tex] follows an arithmetic progression with a difference of 2 between terms, so that[tex]b_n=4+2(n-1)=2n+2[/tex]Then we have[tex]2n+2=a_{n+1}-a_n\implies a_{n+1}=a_n+2n+2[/tex]so that [tex]a_n[/tex] is given recursively by[tex]\begin{cases}a_1=1\\a_{n+1}=a_n+2(n+1)&\text{for }n\ge1\end{cases}[/tex]By substitution, we can try to find a pattern:[tex]a_2=a_1+2\cdot2[/tex][tex]a_3=a_2+2\cdot3=a_1+2(2+3)[/tex][tex]a_4=a_3+2\cdot4=a_1+2(2+3+4)[/tex][tex]a_5=a_4+2\cdot5=a_1+2(2+3+4+5)[/tex]and so on, with the general pattern[tex]a_n=a_1+2(2+3+4+\cdots+n)[/tex]and since [tex]a_1=1[/tex] we can write this as[tex]a_n=1+2(2+3+4+\cdots+n)[/tex][tex]a_n=-1+2+2(2+3+4+\cdots+n)[/tex][tex]a_n=2(1+2+3+4+\cdots+n)-1[/tex]Recall that[tex]\displaystyle\sum_{i=1}^ni=1+2+3+\cdots+n=\dfrac{n(n+1)}2[/tex]Then[tex]a_n=2\dfrac{n(n+1)}2-1\implies\boxed{a_n=n^2+n-1}[/tex]