1 5 12 22 sequence

Asked by maham237 @ in Mathematics viewed by 258 People

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.

Answered by maham237 @


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 nth 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.


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