Question 1014011
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First we'll try a sum of nth powers with the first two terms:
If that doesn't work, we'll try the difference of nth powers 

{{{a[n]}}}{{{""=""}}}{{{x^n+y^n}}}

{{{5}}}{{{""=""}}}{{{x^1+y^1}}}
{{{5}}}{{{""=""}}}{{{x^2+y^2}}}

{{{system(x+y=5,x^2+y^2=5)}}}

{{{matrix(8,1,
y=5-x,
x^2+(5-x)^2=5,
x^2+(25-10x+x^2)=5,
x^2+25-10x+x^2=5,
2x^2-10x+25=5,
2x^2-10x+20=0,
x^2-5x+10=0,
imaginary)}}}

No good. Comes out with imaginary values for x and y

So we try a difference of powers for the first two terms:

{{{a[n]}}}{{{""=""}}}{{{x^n-y^n}}}

{{{5}}}{{{""=""}}}{{{x^1-y^1}}}
{{{5}}}{{{""=""}}}{{{x^2-y^2}}}

{{{system(x-y=5,x^2-y^2=5)}}}

{{{matrix(7,1,
x-5=y,
x^2-(x-5)^2=5,
x^2-(x^2-10x+25)=5,
x^2-x^2+10x-25=5,
10x-25=5,
10x=30,
x=3)}}}

{{{matrix(3,1,
x-5=y,
3-5=y,
-2=y)}}}

So the formula we get from the first two terms,
substituting x=3 and y=-2 in

{{{a[n]}}}{{{""=""}}}{{{x^n-y^n}}}

becomes

{{{a[n]}}}{{{""=""}}}{{{3^n-(-2)^n}}}

But we must check to see if it holds for all the given 
numbers in the series:

{{{a[1]}}}{{{""=""}}}{{{3^1-(-2)^1}}}{{{""=""}}}{{{3-(-2)}}}{{{""=""}}}{{{3+2}}}{{{""=""}}}{{{5}}}

{{{a[2]}}}{{{""=""}}}{{{3^2-(-2)^2}}}{{{""=""}}}{{{9-(4)}}}{{{""=""}}}{{{5}}}

{{{a[3]}}}{{{""=""}}}{{{3^3-(-2)^3}}}{{{""=""}}}{{{27-(-8)}}}{{{""=""}}}{{{27+8}}}{{{""=""}}}{{{35}}}

{{{a[4]}}}{{{""=""}}}{{{3^4-(-2)^4}}}{{{""=""}}}{{{81-(16)}}}{{{""=""}}}{{{65}}}

{{{a[5]}}}{{{""=""}}}{{{3^5-(-2)^5}}}{{{""=""}}}{{{243-(-32)}}}{{{""=""}}}{{{243+32}}}{{{""=""}}}{{{275}}}

So it works for all the given terms, so the formula is:

{{{a[n]}}}{{{""=""}}}{{{3^n-(-2)^n}}}

Edwin</pre>