Hello,

Here are some suggestions:

Find f(n) such that b_n=a_n-f(n) has a nice property.

a-c) a_n-f(n)=a_{n-1}-f(n-1)

d) a_n-f(n)=2(a_{n-1}-f(n-1))

I think we can find f(n) as a polynomial of n.

For example,

a) f(n)-f(n-1)=3(n-1). Let f(n)=pn^2+qn.

By substituting, we see that p=3/2 and q=-3/2 works.

Thus, b_n=b_{n-1}=b_0=1.

a_n=f(n)+b_n=f(n)+b_0=(3/2)n^2-(3/2)n+1.

e) I am not sure what you mean by "2n2",

but the same method should work if this is "2n^2".

a-c) It is also common to consider the sequence obtained by

b_n=a_n-a_{n-1}.

a_n=a_0+(b_1+b_2+...+b_n).

For example,

c)a_n=a_0+3(1^2+2^2+...+n^2)

=10+3(1/6)n(n+1)(2n+1))=10+n(n+1)(2n+1)/2.

I hope it helps.