The following is a Generating function question:

The answer is provided below, but I do not fully understand the answer.

I have highlighted the parts of the answer which I would like a more thorough explanation.

l (i) We write down the following: x): 1+k$+2$2+... (1 an.) (1 :1. - This solva to: _ a: 1+[k1}x \"Il[lx)3{1+x)+[1:c}{1+x}' \" W' l1: (n) It ]_ A B C n flirt_[11:)3+[1s:}2+1m+1+:r we get the formula: [1+e)A+(1m)[1+e]3 + [1 :c}2(1 +s]C+ (1 @313 =e+ [1+ [k 1):c)(1 x]2. Recalling that the general solution for [111)]- is an = (n?) and for is of. = (1)", we discover that A is the leading term and determines the asymptotic behaviour. To determine the value for A, using the cover-up rule, substitute a: = 1 for 2A = 1 or A = 1,!2. In other words. the value of k is irrelevant, the sequences always go to positive innity. (iii) Letting k = 1, we have to determine the values for A,B,C,D according to the formula: (1 + 1:)A+ (1 :c}{1 +$)B + (1 1:)2(1 +m)C + (1 $)3D = s + (1 ::.')2. We know already that A = 1,32 regardless. Plug in m = 1 for D = 3f8. Another good choiceis :r: = U for 1f2+B+C+3f8= 1 and B+C= 1f8.Lastlyt1-3r :1: =2 for: 3 3 533+3C_3 or C B = 5,18. The solution is B = 1f4 and C = 3(3. The formula is: _1 11+? _1 n+1 E E_ n \""'2( n) 4(n)+3+8(1}' Let k be an integer and let ao, a1, a2, ... be the sequence defined by an = an-2 + n for all n 2 2, do = 1 and a1 = k. (i) Determine the generating function for the above sequence do, a1, a2, . . . . (ii) Determine for which integers k, if any, the sequence tends to negative infinity as n tends to positive infinity. iii) Letting k = 1 in the above recurrence relation, find an explicit expression for an in terms of n