Simple sequence made complicated

Take a look here .

By the algorithms of solving homogenous recurrence sequence, we will subsitute $a_{n}$ with $r^{n}$ .

$r^{n} - 4r^{n-1} + 5r^{n-2} - 2r^{n-3} = 0$

$r^{n-3}(r^{3} -4r^{2} + 5r -2) = 0$

Since the value of $r^{n-3}$ will never be 0, we will divide both sides by $r^{n-3}$

$(r^{3} -4r^{2} + 5r -2) = 0$

$(r-1)(r-1)(r-2) = 0$

Thus, the general form of this sequence is

$a_{n} = A(2^{n}) + B(1^{n}) + Cn(1^{n})$

$= A(2^{n}) + B + Cn$

Substituting the values of $a_{0}, a_{1}, a_{2}$ into the general form, we will get our system of equations

$a_{0} = A + B = 1$

$a_{1} = 2A + B + C = 2$

$a_{2} = 4A + B + 2C = 3$

Solving the system of equations,

$A= 0, B = 1, C = 1$

Substituting the values of $A, B, C$ into our general form,

$\boxed{a_{n} = n + 1}$

Thus,

$a_{1000} = \boxed{1001}$

it is an AP series with first term(a)=1, D=1 solve it by putting n=1001 in following wqn. Tn=a+(n-1)D=a_1000

I took a look at the wikipedia article too, and I considered applying the Python sympy library. Then I glanced again at the title of this puzzle and saw the word 'simple'! Once again I reminded myself that even great mathematicians weren't averse to experimentation.

>>> a=[1,2,3]

>>> for k in range(10):

... a.append( 4 a[-1] - 5 a[-2] + 2*a[-3] )

...

>>> a

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

It doesn't take much in this case.