Question

lf $n>3$, then $xyz{C}_0-(x-1)(y-1)(z-1)C_1+(x-2)(y-2)(z-2)C_2$
$-(x-3)(y-3)(z-3)C_3$+.....+$(-1{)}^n(x-n\left)\right(y-n\left)\right(z-n)C_n$ equal to

• A. $xyz$
• B. $nxyz$
• C. $-xyz$
• D. $0$

Answer 1

The correct option is D $0$

The given expression in for of sum can be written as
${\sum }_{r=0}^n(-1{)}^{r}xyz{C}_r+{\sum }_{r=0}^n(-1{)}^{r}r(x+y+z)C_r+{\sum }_{r=0}^n(-1{)}^r{r}^2(xy+yz+zx)C_r-{\sum }_{r=0}^n(-1{)}^r{r}^3\left(xyz\right)C_r$
Since $n>3$, the above terms form alternate positive and negative pairs thus cancelling each other and yielding zero.
Thus the final answer is $0$.