Find the closed form of the generating function A(x) for the following a_n.?

For (a), we have, writing G(x) = G{a(n)} for the generating function of a(n):

G{2^n} = ∑[k = 0,+∞](2^k)(x^k) = 1/(1−2x)

G{3^n} = ∑[k = 0,+∞](3^k)(x^k) = 1/(1−3x)

G{1(n)} = ∑[k = 0,+∞](x^k) = 1/(1−x)

Therefore:

G(x) = 1/(1−2x) + 1/(1−3x) − 1/(1−x)

For (b), we have:

a(0) = 0

a(n + 1) = n + 1 + a(n), n ≥ 0

And G{a(n + 1)} = G{a(n)}/x − a(0) = G(x)/x

Therefore:

G(x)/x = G{n + 1} + G(x) =

= 1/(1−x)^2 + G(x) ⇒

⇒ (1−x)G(x) = x/(1−x)^2 ⇒

⇒ G(x) = x/(1−x)^3

Or use the general result:

If s(n) = ∑[k = 0,n]k, then:

S(x) = G{n}/(1−x)

For (c), note that:

C(n + 4,n) = (n + 4)(n + 3)(n + 2)(n + 1)/4!

And:

G(x) = G{(−2)^n} = ∑[k = 0,+∞](−2x)^k = 1/(1 + 2x)

Therefore:

(x^4)G(x) = ∑[k = 0,+∞](−2x)^(k + 4) = (2x^4)/(1 + 2x)

And:

d/dx[(x^4)G(x)] = ∑[k = 0,+∞](−2)(k + 4)(−2x)^(k + 3) =

= (4x^3)(3x + 2)/(2x + 1)^2

(d/dx)^2[(x^4)G(x)] = ∑[k = 0,+∞]4(k + 4)(k + 3)(−2x)^(k + 2) =

= (8x^2)(6x^2+ 8x + 3)/(2x + 1)^3

(d/dx)^3[(x^4)G(x)] = ∑[k = 0,+∞](−8)(k + 4)(k + 3)(k + 2)(−2x)^(k + 1) =

= 48x(2x^3 + 4x^2 + 3x + 1)/(2x + 1)^4

Finally:

(d/dx)^4[(x^4)G(x)] = ∑[k = 0,+∞]16(k + 4)(k + 3)(k + 2)(k + 1)(−2x)^k =

= 48/(2x + 1)^5

Therefore:

∑[k = 0,+∞][(k + 4)(k + 3)(k + 2)(k + 1)/4!](−2x)^k = 1/[8(2x + 1)^5]

PS: please check calculations.

c_n + 3