1.

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

Answer»

`7`
`8`
`9`
`11`

Solution :`(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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