A) \[\frac{r!{{(1)}^{r}}}{(n-r)!}\]
B) \[\frac{{{(-1)}^{r}}}{r!(n-r)!}\]
C) \[\frac{1}{r!(n-r)!}\]
D) None of these
Correct Answer: B
Solution :
We have, \[\frac{1}{x(x+1)(x+2)...(x+n)}=\frac{{{A}_{0}}}{x}+\frac{{{A}_{1}}}{x+1}+...+\] \[\frac{{{A}_{r}}}{x+r}+...+\frac{{{A}_{n}}}{x+n}\] \[\Rightarrow \]\[1={{A}_{0}}(x+1)(x+2)...(x+n)\] \[+...+{{A}_{r}}x(x+1)(x+2)...(x+r-1)(x+r+1)\] \[...(x+n)+...+{{A}_{n}}x(x+1)...(x+n-1)\] Putting\[x=-r\], we get; \[1={{A}_{r}}(-r)(-r+1)(-r+2)...(-3)(-2)(-1)(1)(2)\] \[...(n-r)\] \[\Rightarrow \,1={{A}_{r}}{{(-1)}^{r}}\{r(r-1)\,(r-2)\,...(3\cdot \,2\cdot 1\}\] \[(1\cdot 2\cdot 3...(n-r))\] \[\Rightarrow \] \[{{A}_{r}}={{(-1)}^{r}}/r!(n-r)!\]
You need to login to perform this action.
You will be redirected in
3 sec
