Question

sum of n terms of 1^3+3^3.........

Answer 1

you are given only two terms of series , can't possible to identify which type of series,
but according to your question, I think
question is
1³ + 3³ + 5³ + 7³ +.......(2n-1)³ = ?
use sigma concept here,
sigma (2r -1)³ where[ r = 1 to n ] = 1³ + 3³ + 5³ + ....(2n-1)³

= sigma{ 8r³ -3.(2r)².1 + 3.(2r).1² - (1)³}
=8 sigma .r³ -12sigma. r² +6 sigma r - sigma

we know,
sigma r³[ 1 ≤ r ≤ n ] = {n(n+1)/2}²
sigma r² [1 ≤ r ≤ n ] = n(n +1)(2n+1)/6
sigma r [1 ≤ r ≤ n ] = n(n +1)/2
sigma 1 [1 ≤ r ≤ n ] = n

use this above .
hence,
Sn = 8.{n(n+1)/2}² -12n(n+1)(2n+1)/6 +6n(n+1)/2 -n