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The sum of the series ` sum_(r=0) ^(n) ""^(2n)C_(r), ` isA. `2^(2n)`B. `2^(n)`C. `2^(2n) + ""^(2n)C_(n)`D. `(1)/(2) (2^(2n) + ""^(2n)C_(n))` |
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Answer» Correct Answer - d We have, `sum_(r=0)^(n) ""^(2n)C_(r) = (1)/(2)sum_(r=0)^(n) (2^(2n) + ""^(2n)C_(n))` ltbbrgt `rArr sum_(r=0)^(n) ""^(2n)C_(r) = (1)/(2)sum_(r=0)^(n) ""^(2n)C_(2n) + ""^(2n)C_(n)r = (1)/(2)[ {sum_(r=0)^(2n) ""^(2n)C_(r)}+""^(2n)C_(n)r]` `rArr sum_(r=0)^(n) ""^(2n)C_(r) = (1)/(2) (2^(2n) + ""^(2n) C_(n))` . |
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