Question: Solve the following recurrence relation by examining the first few values for a formula and then prove your conjectured formula by induction

$h\_n=h\_{n-1} -n+3, \ (n\geq 1); \ h\_0=2$

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Here I've gone all the way to $n=8$ and the only pattern I've noticed is in the differences. By that I mean, 

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$h\_0=2$

$h\_1=2-1+3=4;$ $\ \ \ \ $from $h\_1$ to $h\_0$ there's a difference of $+2$ 

$h\_2=4-2+3=5;$$\ \ \ \ $from $h\_2$ to $h\_1$ there's a difference of $+1$ 

$h\_3=5-3+3=5;$$\ \ \ \ $from $h\_3$ to $h\_2$ there's a difference of $+0$

$h\_4=5-4+3=4;$$\ \ \ \ $from $h\_4$ to $h\_3$ there's a difference of $-1$

$h\_5=4-5+3=2;$$\ \ \ \ $from $h\_5$ to $h\_4$ there's a difference of $-2$

$h\_6=2-6+3=-1;$$\ \ \ \ $from $h\_6$ to $h\_5$ there's a difference of $-3$ 

$h\_7=-1-7+3=-5;$$\ \ \ \ $from $h\_7$ to $h\_6$ there's a difference of $-4$

$h\_8=-5-8+3=-10;$$\ \ \ \ $from $h\_8$ to $h\_7$ there's a difference of $-5$

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I'm not really sure what this means as far as the formula goes though

Note: I'm only struggling with coming up with the formula. I can handle the induction.

Answer: 
\*\*Hint: \*\*Write it in terms of $n$.$$h\_n = h\_{n-1}-n+3,$$
but $$h\_{n-1}=h\_{n-2}-(n-1)+3,$$
so $$h\_n = h\_{n-2}-n-(n-1)+2\*3,$$
replace it now using $h\_{n-2}=h\_{n-3}-(n-2)+3,$ you will get $$h\_n=h\_{n-3}-(n-2)-(n-1)-n+3\*3,$$
Do it again and again and the pattern will pop out.  

So what is $h\_{n-k}?$ what happens when $k=n?$
You have: $h\_n = (h\_n - h\_{n-1}) + (h\_{n-1} - h\_{n-2})+ ...+ (h\_2 - h\_1) + (h\_1-h\_0) + h\_0 = -n+3 - (n-1) + 3 - (n-2) + 3 + ... + (-1 + 3) + 2 = -(1+2++\cdots + n) + 3n + 2 = $? Can you finish it off?