Question

Consider the sequence 1, 5, 12, 22, 35, 51, . . . (with a0 = 1). By looking at the differences between terms, express the sequence as a sequence of partial sums. Then find a closed formula for the sequence by computing the nth partial sum.

Answer 1

Let $a_n$ denote the $n$-th term of the sequence, starting with the index $n=0$. Denote by $b_n$ the forward differences of $\{a_n\}_{n\ge0}$, that is

$b_n=a_{n+1}-a_n$

for $n\ge0$. We have

$\{b_n\}_{n\ge0}=\{4,7,10,13,16,\ldots\}$

which indicates $b_n$ is an arithmetic sequence with common difference between terms of 3 and starting with 4, so that

$b_n=4+3n$

for $n\ge0$. So we have

$a_0=1$

$a_1=a_0+b_0=1+4$

$a_2=a_0+b_0+b_1=1+4+7$

$a_3=a_0+b_0+b_1+b_2=1+4+7+10$

and so on, with the $n$th term given by

$a_n=a_0+b_0+b_1+\cdots+b_{n-2}+b_{n-1}$

$a_n=1+4+7+\cdots+(4+3(n-2))+(4+3(n-1))$

$a_n=1+4+7+\cdots+(3n-2)+(3n+1)$

Reversing the order of all but the first term in the sum gives

$a_n=1+(3n+1)+(3n-2)+(3n-5)+\cdots+7+4$

so that

$2a_n=2+(4+3n+1)+(7+3n-2)+\cdots+(3n-2+7)+(3n+1+4)(3n+1+1)$

$2a_n=2+\underbrace{(3n+5)+(3n+5)+\cdots+(3n+5)+(3n+5)}_{n\text{ times}}$

$2a_n=2+n(3n+5)$

$\implies\boxed{a_n=1+\dfrac{n(3n+5)}2}$

for $n\ge0$.