`(6^(n+3)-32*6^(n+1))/(6^(n+2)-2*6^(n+1))=1`

The value of (6^(n+3) - 32.6^(n+1))/(6^(n+2) - 2.6^(n+1)) is equal to

Find the sum sum_(n=1)^(oo)(6^(n))/((3^(n)-2^(n))(3^(n+1)-2^(n+1)))

Find the sum sum_(n=1)^(oo)(6^(n))/((3^(n)-2^(n))(3^(n+1)-2^(n+1)))

If (6^(n))/(6^(-2))=6^(3) , then n=?

Using Mathematical induction prove that 3.2^(2)+3^(2).2^(3)+3^(3).2^(4)+…………+3^(n).2^(n+1)=12/6(6^(n)-1) , for all n in N

The value of quad 1^(6)+2^(6)+3^(6)...n^(6)lim_(n rarr oo)(1^(6)+2^(6)+3^(6)...n^(6))/((1^(2)+2^(2)+3^(2)+...n^(2))(1^(3)+2^(3)+...n^(3)))

If n is a non zero rational number then show that 1 + n/2 + (n (n - 1))/(2.4) + (n(n-1)(n - 2))/(2.4.6) + ….. = 1 + n/3 + (n (n + 1))/(3.6) + (n (n + 1) (n + 2))/(3.6.9) + ….

If n is a non zero rational number then show that 1 + n/2 + (n (n - 1))/(2.4) + (n(n-1)(n - 2))/(2.4.6) + ….. = 1 + n/3 + (n (n + 1))/(3.6) + (n (n + 1) (n + 2))/(3.6.9) + ….

1+ (n)/(3)+(n (n+1))/(3.6)+(n (n+1) (n+2))/(3.6.9)+... oo