Question

If n is an odd natural number, then `sum_(r=0)^n (-1)^r/(nC_r)` is equal to

Answer 1

Correct Answer - a
We have,
`sum_(r=0)^(n) ((-1)^(r))/(""^(n)C_(r) )= sum_(r=0)^((n+1)/(2)) {(( -1)^(r))/(""^(n)C_(r) )+ (-1)^(n-r)/(""^(n)C_(n-r))}`
`rArr sum_(r=0)^(n) ((-1)^(r))/(""^(n)C_(r) )= sum_(r=0)^((n+1)/(2))(-1)^(r) {(( -1)^(r))/(""^(n)C_(r) )+ (-1)^(n-r)/(""^(n)C_(n-r))}`
`rArr sum_(r=0)^(n) ((-1)^(r))/(""^(n)C_(r) )= sum_(r=0)^((n+1)/(2))(-1)^(2) {( 1)/(""^(n)C_(r) )+ (1)/(""^(n)C_(r))}= 0 [{:(therefore " is odd and "),(""^(n)C_(r) = ""^(n)C_(n-r)):}]` . s