if `6n` tickets numbered 0,1,2,....... 6n-1 are placed in a bag and three are drawn out , show that the chance that the sum of the numbers on then is equal to 6n is `(3n)/((6n-1)(6n-2))`

if 6n tickets numbered 0126n-1 are placed in a bag and three are drawn out,show that the chance that the sum of the numbers on them is equal to 6n is (3n)/((6n-1)(6n-2))

Find the sum to n terms of the series whose nth term is 4n^(3)+6n^(2)+2n

Prove by mathematical induction that the sum of the squares of first n natural numbers is 1/6n(n+1)(2n+1) .

Statement 1: Out of 5 tickets consecutively numbered, three are drawn at random. The chance that the numbers on them are in A.P. is 2//15 . Statement 2: Out of 2n+1 tickets consecutively numbered, three are drawn at random, the chance that the numbers on them are in A.P. is 3n//(4n^2-1) .

Statement 1: Our of 5 tickets consecutively numbered, three are drawn at random. The chance that the numbers on them are in A.P. is 2//15 . Statement 2: Out of 2n+1 tickets consecutively numbered, three are drawn at random, the chance that the numbers on them are in A.P. is 3n//(4n^2-1) .

Statement 1: Our of 5tickets consecutively numbered,three are drawn at random.The chance that the numbers on them are in A.P.is 2/15. Statement 2: Out of 2n+1 tickets consecutively numbed,three are drawn at random,the chance that the numbers on them are in A.P.is 3n/(4n^(2)-1) .

Solve : (2n + 3)/(6n -5) = 1

The sum of the series sum_(n=1)^oo(n^2+6n+10)/((2n+1)!) is equal to

Using the mathematical induction, show that for any natural number n, 1/2.5 + 1/5.8 + 1/8.11 + …+ 1/((3n-1)(3n+2))=n/(6n+4)

If the numbers (2n-1), (3n+2) and (6n-1) are in AP, find n and hence find these numbers.