The sum of the first $n$ terms of the series ${1^2} + {2.2^2} + {3^2} + {2.4^2} + {5^2} + {2.6^2} + \;...$ is $\frac{{n{{\left( {n + 1} \right)}^2}}}{2}$ where n is even. When $n$ is odd the sum is

$\frac{{3n\left( {n + 1} \right)}}{2}$

$\frac{{n{{\left( {n + 1} \right)}^2}}}{4}$

${\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2}$

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Analyzing +4 | -1

The sum of the first

n

terms of the series

{1^2} + {2.2^2} + {3^2} + {2.4^2} + {5^2} + {2.6^2} + \;...

\frac{{n{{\left( {n + 1} \right)}^2}}}{2}

where n is even. When

n

is odd the sum is

View answer and explanation

Correct option: B

\frac{{{n^2}\left( {n + 1} \right)}}{2}

Option B (

\frac{{{n^2}\left( {n + 1} \right)}}{2}

) is correct because it represents the sum of the series when n is odd, derived by adding the (2m+1)th term to the sum of 2m terms.