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