Find the sum of n terms of the series `1^3+3.2^2+3^3+3.4^2+5^3+3.6^2+.......` when (i)n is odd (ii)n is even

Explanation

Case I If n is even.
Let `n=2m`
`:. S=1^(3)+3*2^(2)+3^(3)+3*4^(2)+5^(3)+3*6^(2)+"......"+(2m-1)^(3)+2(2m)^(2)`
`={1^(3)+3^(3)+5^(3)+"......"+(2m-1)^(3)}+3{2^(2)+4^(2)+6^(2)+"....."+(2m)^(2)}`
`=sum_(r=1)^(m)(2r-1)^(3)+3*4sum_(r=1)^(m)r^(2)`
`=sum_(r=1)^(m){8r^(3)-12r^(2)+6r-1}+12sum_(r=1)^(m)r^(2)`
`=8sum_(r=1)^(m)r^(3)-12sum_(r=1)^(m)r^(2)+6sum_(r=1)^(m)r-sum_(r=1)^(m)1+12sum_(r=1)^(m)r^(2)`
`=8sum_(r=1)^(m)r^(3)+6sum_(r=1)^(m)r-sum_(r=1)^(m)1`
`=8*(m^(2)(m+1)^(2))/(4)+6(m(m+1))/(2)-m`
`=m[2m^(3)+4m^(2)+5m+2]`
`=(n)/(2)[2((n)/(2))^(3)+4((n)/(2))^(2)+5((n)/(2))+2] " "[:.m=(n)/(2)]`
Hence, `S=(n)/(2)(n^(3)+4n^(2)+10n+8)" " ".....(i)"`
Case II If n is odd.
Then, `(n+1)` is even in the case
Sum of n terms = Sum of first `(n+1)` terms `-(n+1)th` term
`=((n+1))/(8)[(n+1)^(3)+4(n+1)^(2)+10(n+1)+8]-3(n+1)^(2)`
, `=((n+1))/(8)[(n+1)^(3)+4(n+1)^(2)+10(n+1)+8]-3(n+1)^(2)`
Hence, `S=(1)/(8)(n+1)[n^(3)+7n^(2)-3n-1]`.