Question 893471
<pre>

x| 1| 2| 3| 4| 5| 6| 7|
-|--|--|--|--|--|--|--|
y| 2| 5| 6|11|12|19|20|

(1,2), (2,5), (3,6), (4,11), (5,12), (6,19), (7,20)

Taking the points where x is odd:

We find the quadratic equation through 

(1,2), (3,6), (5,12), (7,20)  

by substituting three of those points in y = ax˛+bx+c

{{{y = (x^2+4x+3)/4}}}

Similar we take the points where x is even and find the
quadratic equation through 

(2,5), (4,11), (6,19) where x is even

{{{y = (x^2+6x+4)/4}}}

We compare the two quadratic equations:

When x is odd the coefficient of x in the parentheses is 4, and
when x is even the coefficient of x in the parentheses is 6

The average of 4 and 6 is 5

4 = 5-1 and 6 = 5+1

So the coefficient of x is 5+(-1)<sup>x</sup> in either case

---

When x is odd, the constant term in the parentheses is 3, and
when x is even, the constant term in the parentheses is 4

The average of 3 and 4 is 7/2

3 = 7/2-1/2 and 4 = 7/2+1/2

the constant term in the parentheses is 7/2+1/2(-1)<sup>x</sup> in either case

So the general term is

{{{y = (x^2+(5+(-1)^x)x+7/2+(1/2)(-1)^x)/4}}}

which simplifies to

{{{y=(2x^2+(2x+1)(-1)^x+10x+7)/8}}}

In sequence notation we replace y by a<sub>n</sub> and x by n

{{{a[n]=(2n^2+(2n+1)(-1)^n+10n+7)/8}}}

Edwin</pre>