If `(1+x)^n=C_0+C_1x+C_2 x^2+...+C_n x^n` , `n in N` ,then `C_0-C_1+C_2-.....+(-1)^(n-1)C_(m-1),` is equal to `(mltn)`

Options

A. A. `((n-1)(n-2)"….."(n-m+1))/((m-1)!) (-1)^(m-1)`
B. B. `.^(n-1)C_(m-1)`
C. C. `((n-1)(n-2)"….."(n-m))/((m-1)!)(-1)^(m-1)`
D. D `.^(n-1)C_(n-m)(-1)^(m-1)`

Answer

A, B, D

Explanation

`((n-1)(n-2)"……"(n-m+1))/((m-1)!)`
= `=((n-1)(n-2)"...."(n-m+1)(n-m)"...."2.1)/((n-m)!(m-1)!)`
`=.^(n-1)C_(m-1)`
`=` Coefficient of `x^(m-1)` in `(1+x)^(n-1)`
`=` Coefficient of `x^(m-1)` in `(1+x)^(n)(1+x)^(-1)`
Now,
`(1+x)^(n) = C_(0)+C_(1)x+C_(2)x^(2)+"...."+C_(m-1)x^(m-1)+"...."+C_(n)x^(n)" "(1)`
`(1+x)^(-1) = 1-x+x^(2)-x^(3)+"...."+(-1)^(m-1)x^(m-1)+"......"" "(2)`
Collecting the coefficients of `x^(m-1)` in the product of (1) and (2), we get
`(-1)^(m-1)C_(0)+(-1)^(m-2)C_(1)+"....."+C_(m-1)`
`=` Coefficients of `x^(m-1)` in `(1+x)^(n-1)`
`=.^(n-1)C_(m-1)`
`:. C_(0) - C_(1)+C_(2)-"......"+(-1)^(m-1)C_(m-1)`
`= .^(n-1)C_(m-1)(-1)^(m-1)`
`= ((n-1)(n-2)"...."(n-m+1))/((m-1)!)(-1)^(m-1)`