Question

The sum of the series `C 20()_0-C 20()_1+C 20()_2-C 20()_3+...-...+C 20()_(10)` is: (1) `-C 20()_(10)` (2) `1/2C 20()_(10)` (3) 0 (4) `C 20()_(10)`

Answer 1

Correct Answer - d
We have,
`""^(20)C_(0)-""^(20)C_(1)+ ""^(20)C_(2)-""^(20)C_(3)+...+ ""^(20)C_(10)-""^(20)C_(11) = 0`
`rArr 2(""^(20)C_(0)-""^(20)C_(1)+ ""^(20)C_(2)-""^(20)C_(3)+...+ ""^(20)C_(9)-""^(20)C_(10) ) - ""^(20)C_(10) = 0`
` [ because ""^(n)C_(r) = ""(n)C_(n-r)]`
`rArr ""^(20)C_(0)-""^(20)C_(1)+ ""^(20)C_(2)-""^(20)C_(3)+...+ ""^(20)C_(9)-""^(20)C_(10) = (1)/(2) ""^(20)C_(10)` .