Sum of series `t_n` = `(n+1) (n+2) (n+3)`

Sum of the series S_(n) =(n) (n) + (n-1) (n+1) + (n-2) (n+2) + …+ 1(2n-1) is

The absolute value of the sum of first 20 terms of series, if S_(n)=(n+1)/(2) and (T_(n-1))/(T_(n))=(1)/(n^(2))-1 , where n is odd, given S_(n) and T_(n) denotes sum of first n terms and n^(th) terms of the series

Find the sum of series upto n terms ((2n+1)/(2n-1))+3((2n+1)/(2n-1))^(2)+5((2n+1)/(2n-1))^(3)+

In a series, if t_n = (n^(2) -1)/(n+1) , then S_6 - S_3= _______.

if in a series t_(n)=(n+1)/((n+2)!) then sum_(n=0)^(10)t_(n) equal to