<s>[INST] Calculate $(2013)^4 - 4(2011)^4+6(2009)^4-4(2007)^4+(2005)^4 = $ Calculate $(2013)^4 - 4(2011)^4+6(2009)^4-4(2007)^4+(2005)^4 = $
Try:: Let $x = 2009$, Then expression convert into $(x+4)^4-4(x+2)^4+6x^4-4(x-2)^4+(x-4)^4$
$\left\{(x+4)^4+(x-4)^4\right\}-4\left\{(x+2)^4+(x-2)^4\right\}+6x^4$
But This is very Complicated for calculation.
can anyone explain me  better idea for that expression
If yes plz explain me
Thanks
 [/INST] Hint: This is a finite differences problem. 
If $x(n)$ is any function, then $$(\Delta^4 x)(n)= x(n+4)-4x(n+3)+6x(n+2)-4x(n+1)+x(n)$$
Where $\Delta$ is the finite difference operator.
Now, since $x(n)=(2n+1)^4$, is a polynomial of degree $4$ with lead coefficient $2^4$ we know that $(\Delta^4x)(n)$ is constant, equal to $2^4\cdot 4!$.
Generalizing with $5$ instead of $4$, for example:
$$2013^5-5\cdot 2011^5 + 10\cdot 2009^5-10\cdot 2007^5 + 5\cdot 2005^5 -2003^5= 2^5\cdot 5!$$
or:
$$2013^1-2011^1 = 2^1\cdot 1!$$
$$2013^2-2\cdot 2011^2 + 2009^2 = 2^2\cdot 2!$$
$$2013^3-3\cdot 2011^3+3\cdot 2009^3 - 2007^3 = 2^3\cdot 3!$$
</s>