The value of $\triangle^5[(1-x)(1-x^2)(1+x^2)]$ is
(a) 5!
(b) 0
(c) 1
(d) 5
I am not getting how to deal with this sum.. What I know is $\triangle f(x)=f(x+1)-f(x)$, and if I use this to obtain $\triangle^5[(1-x)(1-x^2)(1+x^2)]$, then it will be very complicated to solve..
So instead of directly solving this I tried to obtained $\triangle^5 f(x)$ as formula
So I got
$$\triangle^2 f(x)=f(x+2)-2f(x+1)+f(x)$$
$$\triangle^3 f(x)=f(x+3)-3f(x+2)+3f(x+1)-f(x)$$
$$\triangle^4 f(x)=f(x+4)-4f(x+3)+6f(x+2)-4f(x+1)+f(x) $$
$$\triangle^5 f(x)=f(x+5)-5f(x+4)+10f(x+3)-10f(x+2)+4f(x+1)-f(x)$$, clearly its forms binomial expansion type formula..
But still if I susbtitute the values of $f(x+5),f(x+4),f(x+3),f(x+2),f(x+1)$ and $f(x)$, it will be complicate to solve as much multiplications will be involved...
Is there a better and less time taking way, such questions are generally part of a competitive exam... where maximum 2-3 minutes can be given to such questions...
thanks in advance!!
Note that the function simplifies to $(1-x)(1-x^4) = 1 - x -x^4 + x^5$.
Recall that $\Delta x^n = (x+1)^n - x^n$.
So just iterate on that $5$ times. But we see that $\Delta^n x^m = 0$ if $m <n$ (e.g. $\Delta^3 x^2 = 0$). This is because $\Delta x^n$ is a polynomial of degree $n-1$. Hence $n$ differences turns it constant, and $n+1$ will make it zero.
So we can already see $$ \Delta^5 \Big( 1 - x -x^4 + x^5 \Big) = \Delta^5 x^5 $$ with no real computation. What remains is trivial, even if we work from the definition and forego formulas like $\Delta^n x^n = n!$.