Given series is: 

S_n = 5 + 11 + 19 + 29 + 41 +........            ................1

S_n =        5 + 11 + 19 + 29 + 41 +........     ................2

Subtract it, we get

      S_n - S_n = 5 + (11 - 5) + (19 - 11) + (29 - 19) + (41 - 29) + ........... - a_n-1

=> 0 = 5 + 6 + 8 + 10 + .............. - a_n-1

=> a_n-1 = 5 + 6 + 8 + 10 + .............

=> a_n-1 = 5 + {6 + 8 + 10 + .............(n-1)th term}

=> a_n-1 = 5 + [(n-1)/2 *{2*6 + (n-1-1)2}]

=> a_n-1 = 5 + [(n-1)/2 *{12 + (n - 2)2}]

=> a_n-1 = 5 + [(n-1) *{6 + (n - 2)}]

=> a_n-1 = 5 + [(n-1) *(n + 4)]

=> a_n-1 = 5 + (n² + 3n - 4)

=> a_n-1 = n² + 3n + 1

Now, sum of n terms

     S_n = ∑ a_n-1

=> S_n = ∑ (n² + 3n + 1)

=> S_n = ∑n² + 3∑n + ∑1

=> S_n = {n*(n+1)*(2n+1)}/6 + 3*n(n+1)/2 + n

=> S_n = n[{(n+1)*(2n+1)}/6 + 3(n+1)/2 + 1] 

=> S_n = n[(n+1)*(2n+1) + 9(n+1) + 6]/6

=> S_n = n[2n² + 3n + 1 + 9n + 9 + 6]/6

=> S_n = n[2n² + 12n + 16]/6

=> S_n = n[n² + 6n + 8]/3

=> S_n = n*(n+2)*(n+4)/3

So, the sum of n terms = n*(n+2)*(n+4)/3