f(n)=sum_(r=1)^(n)[r^(2)(^(n)C_(r)-^(n)C_(r-1))+(2r+1)(^(n)C_(r))]," then "

If f(x)=sum_(r=1)^(n) { r^(2) (""^(n)C_(r)- ^(n) C_(r-1))+ (2r+1) ^(n) C_(r)} and f(30)=30(2)^(lambda), then the value of lambda is

If f(x)=sum_(r=1)^(n) { r^(2) (""^(n)C_(r)- ^(n) C_(r-1))+ (2r+1) ^(n) C_(r)} and f(30)=30(2)^(lambda), then the value of lambda is

sum_(r=1)^(n)(1)/((r+1)(r+2))*^(n+3)C_(r)=

Find the sum sum_(r=1)^(n)r^(2)(^nC_(r))/(n_(C_(r-1)))

If a_(n)=sum_(r=0)^(n)(1)/(""^(n)C_(r)) then sum_(r=0)^(n)(r)/(""^(n)C_(r)) equal to

If n is a positive integer, prove that sum_(r=1)^(n)r^(3)((""^(n)C_(r))/(""^(n)C_(r-1)))^(2)=((n)(n+1)^(2)(n+2))/(12)

Let f(n)=sum_(r=1)^(n)r^(2)C_(r)^(2)* Then ,f(5)=

Find the sum sum_(r=1)^(n) r^(2) (""^(n)C_(r))/(""^(n)C_(r-1)) .