Find the sum of first `n` terms of the series `1^3+3xx2^2+3^3+3xx4^2+5^3+3xx6^2+ w h e n` `n` is even `n` is odd

Explanation

(i) n is even. Let n =2m. Then
`S_(n)=S_(2m)=sum_(r=1)^(m)(2r-1)^(3)+3sum_(r=1)^(m)(2r)^(2)`
`=sum_(r=1)^(m)[8r^(3)-3(2r)^(2)+3(2r)-1)]+12sum_(r=1)^(m)r^(2)`
`=8sum_(r=1)^(m)r^(3)+6sum_(r=1)^(m)r-sum_(r=1)^(m)1`
`1m^(2)(m+1)^(2)+3m(m+1)-m`
`=m[2m^(3)+4m^(2)+5m+2]`
Put 2m=n or m=n/2
`thereforeS_(n)=n/8[n^(3)+4n^(2)+10n+8]` (1)
(ii) If n is odd,then n + 1 is even. Now,
`S_(n)-S_(n+1)-T_(n+1)` (2)
`S_(n+1)` is obtained from (1) by replacing n by n+1 and `T_(n+1)=(n+1)th` term=`3(n+1)^(2)`. Hence from (2),
`S_(n)` (n odd)
`=(n+1)/8[(n+1)^(3)+4(n+1)^(2)+10(n+1)+8]-3(n+1)^(2)`
`S_(n)=(n+1)/8[n^(3)+7n^(2)-3n-1]`
Equations (1) and (3) give the required results.