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

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

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1

The sum of first

n

terms of the series

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

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

when

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 when n is odd, the last term is n² and the sum can be calculated by adding n² to the sum of the previous (n-1) terms.

Practice All Chapter MCQs →

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{A} \cup({B} \cap {C})

A-(B \cap C)

(B \cup C)-A

a,b,c

a,x,b

b,y,c

{x^2},{b^2},{y^2}

P(x)=1-x+x^2-x^3+\ldots+x^{2012}

y

Q(y)=a_0+a_1 y+\ldots+a_{2012} y^{2012}

y=x-2

\sum\limits_{{i=0}}^{2012} a_i

\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)

\log \left(\frac{a}{3 b}\right)

A P

a, b

G P

a, b, c

\frac{3}{4}