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

Let's calculate the coefficients at powers of $x$.
First, it's clear the coefficient at $x^4$ is zero ($1+1-4-4+6=0$).
Next, from your grouping of terms it's obvious that coefficients at $x^3$ and $x$ are zero, too. Or from another point of view, our function is even.
Then again, the coefficient at $x^2$ is $2\cdot  6\cdot 4^2 - 2\cdot 4\cdot 6\cdot 2^2=0$.
Finally, we have the coefficient at $x^0$, which is $2\cdot 4^4-4\cdot 2\cdot2^4 = 3\cdot 128 = 384$.
