If `a_n=sum_(r=0)^n1/(^n C_r)` , then `sum_(r=0)^n r/(^n C_r)` equals `(n-1)a_n` b. `n a_n` c. `(1//2)n a_n` d. none of these

Options

A. `(n -1)a_(n)`
B. `n a_(n)`
C. `(n)/(2) a_(n)`
D. none of these

Answer

c

Explanation

We have following cases:
In this case, we have
`a_(n)= sum_(r=0)^(n)(1)/(""^(n)C_(r)) = sum_(r=0)^(n//2-1)(1)/(""^(n)C_(r) )+(1)/(""^(n)C_(n-r))`
`rArr sum_(r=0)^(n//2-1)(2)/(""^(n)C_(r) )=2sum_(r=0)^(n//2-1)(1)/(""^(n)C_(r))` ... (i)
`therefore sum_(r=0)^(n) (r)/(""^(n)C_(r) = n sum_(r=0)^(n//2-1)(1)/(""^(n)C_(r))= n. ((a_(n))/(2)) = (n)/(2) a_(n)` [Using (i)]
CASEII When n is even
In this case, we have
`a_(n) = sun_(r=0)^(n) (1)/(""^(n)C_(r)`
`rArr a_(n) = {sum_(r=0)^(n//2-1)((1)/(""^(n)C_(r))+(1)/(""^(n)C_(n-r)))}+(1)/(""^n)C_(n//2)`
`rArr a_(n) = (sum_(r=0)^(n//2-1)(2)/(""^(n)C_(r)))+(1)/(""^(n)C_(n//2))`
`rArr a_(n)=2(sum_(r=0)^(n//2-1)(1)/(""^(n)C_(r)))+(1)/(""^(n)C_(n//2))` ...(ii)
`therefore sum_(r=0)^(n)(r)/(""^(n)C_(r))= sum_(r=0)^(n//2-1)((r)/(""^(n)C_(r))+(n-r)/(""^(n)C_(r)))+ (n//2)/(""^(n)C_(n//2))`
`rArr sum_(r=0)^(n)(r)/(""^(n)C_(r))= (sum_(r=0)^(n//2-1)(n)/(""^(n)C_(r)))+(n//2)/(""^(n)C_(n//2))`
`rArr sum_(r=0)^(n)(r)/(""^(n)C_(r))=(n)/(2) {2 (sum_(r=0)^(n//2-1)(1)/(""^(n)C_(r)))+(1)/(""^(n)C_(n//2))}= (n)/(2) . a_(n)` [ Using (ii)]
Hence, `sum_(r=0)^(n) (r)/(""^(n)C_(r) = (n)/(2) a_(n)` , for all n `in` N.