(3^3-2^3)+(5^3-4^3)+... n the brackets find the sum of series

• sigma[(2n+1)^3-(2n)^3]
  sigma[12*n^2+6*n+1]
  12*sigma(n^2)+6*sigma(n)+sigma(1) ..(sigma i.e, summation is goes upto n=1 to n)
  12*{(n(n+1)(2n+1))/6}+6*{(n(n+1))/2}+n
  n(n+1){4n+2+3}+n
  n[(n+1)(4n+5)+n)
  4n^3+9n^2+6n... (answer)
• 10 years agoHelpfull: Yes(2) No(0)
• ((3*3*3)-(2*2*2))=19=((3+2)^2-3*2)
  ((5*5*5)-(4*4*4))=61=((5+4)^2-5*4)
  for n=summation((n+(n-1))^2-(n*n-1)
  where n starts from 3
  sum=summmation(3n^2-4n+2)//by simplifying above equation
  
  
• 10 years agoHelpfull: Yes(0) No(1)