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
 A: 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:
$$2013^1-2011^1 = 2^1\cdot 1!$$
$$2013^2-2\cdot 2011^2 + 2009^2 = 2^2\cdot 2!$$
$$2013^3-3\cdot 2011^3+3\cdot 2009^3 - 2007^3 = 2^3\cdot 3!$$
A: 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 $2\cdot 4^4-4\cdot 2\cdot2^4 = 3\cdot 128 = 384$.
A: You correctly realised that the binomial coefficients with alternating signs are a key to solving this problem easily. Indeed these are a signature of powers of finite difference operator $\Delta$ operating on sequences of numbers. Writing any sequence in the form $a=(a_i)_{i\in\Bbb N}$ it is defined by $\Delta(a)=(a_{i+1}-a_i)_{i\in\Bbb N}$ (stated differently, $\Delta(a)_i=a_{i+1}-a_i$ for all $i$). Iterating $\Delta$ gives $\Delta^n(a)=\bigl(\sum_k(-1)^{n-k}\binom nka_{i+k}\bigr)_{i\in\Bbb N}$, which can be proved by induction (or by a slick use of the binomial theorem applied to commuting operators).
Now you want to know $\sum_k(-1)^{4-k}\binom 4ka_k=\Delta^4(a)_0$ where $a=(a_k)_{k\in\Bbb N}=\bigl((2013-2k)^4\bigr)_{k\in\Bbb N}$. Now you can easily expand $a_k=(2013-2k)^4$ as polynomial in$~k$ of degree$~4$, and observe that for sequences whose terms are a polynomial of the index, application of $\Delta$ lower gives another such sequence but with degree one less; it follows that here $\Delta^4(a)$ is a constant sequence, and it suffices to know the value of the constant. The same argument shows that the contributions of degree${}<4$ in the expression for $a_k$ are killed by $\Delta^4$, so that $\Delta^4(a)=\Delta^4\bigl((16k^4)_{k\in\Bbb N}\bigr)$. As it is a general fact that $\Delta^n\bigl((k^n)_{k\in\Bbb N}\bigr)$ is the constant sequence with value $n!$ (just like the $n$-th derivative of $x\mapsto x^n$ is a constant function with value $n!$), we get $\Delta^4\bigl((16k^4)_{k\in\Bbb N}\bigr)_0=16\times4!=384$.
