Find the sum of the following series to n terms:

1³ + 3³ + 5³ + 7³ + ……..

Solution:

Let the nth term of the series be given by T_n

We have:

$ {{T}_{n}}~=\text{ }{{\left[ 1\text{ }+\text{ }\left( n-1 \right)2 \right]}^{3}} $

$ =\text{ }{{\left( 2n-1 \right)}^{3}} $

$ =\text{ }{{\left( 2n \right)}^{3}}-3\text{ }{{\left( 2n \right)}^{2}}.\text{ }1\text{ }+\text{ }{{3.1}^{2}}.2n-{{1}^{3~}} $

$ \left[ Since,\text{ }{{\left( a-b \right)}^{3}}~=\text{ }{{a}^{3}}-3{{a}^{2}}b\text{ }+\text{ }3a{{b}^{2}}-b \right] $

$ =\text{ }8{{n}^{3}}-12{{n}^{2}}~+\text{ }6n-1 $

Now, let the sum of n terms of the given series be given by Sn

We can write:

By simplifying further we get,

$ =\text{ }2{{n}^{2}}~{{\left( n\text{ }+\text{ }1 \right)}^{2}}-n-2n\text{ }\left( n\text{ }+\text{ }1 \right)\text{ }\left( 2n\text{ }+\text{ }1 \right)\text{ }+\text{ }3n\text{ }\left( n\text{ }+\text{ }1 \right) $

$ =\text{ }n\text{ }\left( n\text{ }+\text{ }1 \right)\text{ }\left[ 2n\text{ }\left( n\text{ }+\text{ }1 \right)-2\text{ }\left( 2n\text{ }+\text{ }1 \right)\text{ }+\text{ }3 \right]-n $

$ =\text{ }n\text{ }\left( n\text{ }+\text{ }1 \right)\text{ }\left[ 2{{n}^{2}}-2n\text{ }+\text{ }1 \right]-n $

$ =\text{ }n\text{ }\left[ 2{{n}^{3}}-2{{n}^{2}}~+\text{ }n\text{ }+\text{ }2{{n}^{2}}-2n\text{ }+\text{ }1-1 \right] $

$ =\text{ }n\text{ }\left[ 2{{n}^{3}}-n \right] $

$ =\text{ }{{n}^{2}}~\left[ 2{{n}^{2}}-1 \right] $

∴ The sum of the series is n² [2n² – 1]