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If "^(n)C(0)-^(n)C(1)+^(n)C(2)-^(n)C(3)+...

If `"^(n)C_(0)-^(n)C_(1)+^(n)C_(2)-^(n)C_(3)+...+(-1)^(r )*^(n)C_(r )=28` , then `n` is equal to ……

A

`7`

B

`8`

C

`9`

D

`11`

Text Solution

Verified by Experts

The correct Answer is:
C
`(c )` `'^(n)C_(0)-^(n)C_(1)+^(n)C_(2)-^(n)C_(3)+....+(-1)^(r )*^(n)C_(r )`
`="coefficient of" x^(r ) "in the expansion of" (1-x)^(n)(1+x+x^(2)+….)`
`="coefficient of" x^(r ) "in the expansion of" (1-x)^(n)(1-x)^(-1)`
`="coefficient of" x^(r ) "in the expansion of" (1-x)^(n-1)`
`=(-1)^(r )*^(n-1)C_(r )`
`implies(-1)^(r )*^(n-1)C_(r )=28impliesr` must be even
`'^(n-1)C_(r )=28implies^(n-1)C_(r )=7xx4=(7xx8)/(2)=^(8)C_(2)=n-1=8impliesn=9`
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