InterviewSolution
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InterviewSolution
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Consider the expansion of `(1 + x)^(2n+1)` The average of the coefficients of the two middle terms in the expansion isA. `.^(2n+1)C_(n+2)`B. `.^(2n+1)C_(n)`C. `.^(2n+1)C_(n-1)`D. `.^(2n)C_(n+1)` |
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Answer» Correct Answer - B Total no. of terms in the expansion is 2n + 2. The middle two terms will be `n^(th),(n+1)^(th)` term. So. Average = `(.^((2n+1))C_(n)+.^((2n+1))C_(n+1))/(2)` `=[((2n+1)!)/(n!(n+1)!)+((2n+1)!)/((n+1)!n!)]//2` `=((2n+1)!)/(n!(n+1)!)=.^((2n+1))C_(n)` |
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