(2n)/(5)

sum_(n=1)^(n)(4+3r)=(1)/(2)(n-5)(3n+26)

lim_(n to oo)(1-2n^(2))/(5n^(2)-n+100)=

4n+3,(3n^(2)+5),(2n-4)/(5) which of the following cannot be general term of an AP

Find underset(n to oo)lim ((2n^(3))/(2n^(2)+3)+(1-5n^(2))/(5n+1))

If n integers taken at random are multiplied together,then the probability that the last digit of the product is 1,3,7, or 9 is a.2^(n)/5^(n) b.4^(n)-2^(n)/5^(n) c.4^(n)/5^(n) d.none of these

For r=0,1,2---n then 2.C_(0)+7.C_(1)+12.C_(2)+----+(5n+2)*C_(n)= (5n+2)2^(n-1) (5n+3)2^(n) (5n+3)2^(n+1) (5n+4)2^(n-1)

The sum of the digits in (10^(2n^(2)+5n+1)+1)^(2) (where n is a positive integer),is