Consider the sequence 015122235517092Find both a recurrence

Consider the sequence 0,1,5,12,22,35,51,70,92...Find both a recurrence relation and a closed for this sequence

Given that the sequenc 0,1,5,12,22,35,51,70,92,------

First let us assume that the first term a(0)=0

The difference between terms goes as follows +1,+4,+7,+10-----

If we think of that for n=1,2,3,----

We have 1+3(n-1)

=3n-3+1

=3n-2

So the recursive form is a(n)=a(n-1)+3n-2, where a(0)=0

For the closed form, it will be of the form

a(n)=an^2+bn+c------------------>1

For n=0 and solve for c

substitute n=0 in equation 1 we get

a(0)=a(0^2)+b(0)+c

=a(0)+b(0)+c

0=0+0+c

c=0

Now, from equation 1, we get a(n)=a(n^2)+bn+0 [c=0]

=a(n^2)+bn

For n=1 in equation 1, we get

a(1)=a(1^2)+b(1)+c

=a(1)+b(1)+0

a(1)=a+b=1------------->2

For n=2 in equation 1, we get

a(2)=a(2^2)+b(2)+c

=4a+2b+0

5=4a+2b---------------->3

Now Multiplying the equation 2 with 2 then we get 2a+2b=2--------->4

subtract equation 4 from 3 we get (4a+2b)-(2a+2b)=5-2

=>(4a-2a)+(2b-2b)=3

=>2a+0=3

=>a=3/2

now substitute a=3/2 in the equation a+b=1

=>(3/2)+b=1

=>b=1-(3/2)

=>b=(2-3)/2=-1/2

therefor b=-1/2

Finally a(n)=a(n^2)+bn

=(3/2)(n^2)-(1/2)n

=(3(n^2)/2)-(n/2)

=(1/2)(3(n^2)-n)

which is the closed form of the given sequence.