InterviewSolution
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InterviewSolution
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What is ` sum_(r=0)^(1) ""^(n+r)C_(n)` equal to ?A. `.^(n+2)C_(1)`B. `.^(n+2)C_(n)`C. `.^(n+3)C_(n)`D. `.^(n+2)C_(n+1)` |
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Answer» Correct Answer - A::D `underset(r=0)overset(1)sum .^(n+r)C_(n)=.^(n)C_(n)+.^(n+1)C_(n)` `=1+((n+1)!)/((n+1-n)!n!)=1+((n+1)(n!))/(n!)` `=1+n+1=n+2` `.^(n+2)C_(n+1)((n+2)!)/((n+2-n-1)!(n+1)!)` `=((n+2)(n+1)!)/((n+1)!)=n+2` `" "`OR `underset(r=0)overset(1)sum .^(n+r)C_(n)=.^(n)C_(n)+.^(n+1)C_(n)` `=1+(n+1)=n+2` Now, `.^(n+2)C_(1)=((n+2)!)/(1!(n+2-1)!)=((n+2)(n+1)!)/((n+1)!)=(n+2)` `therefore` Option (a and d) is correct. |
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