The sum to $n$ terms of the series ${2^2} + {4^2} + {6^2} + ...$ is

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

$\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$

$\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{9}$

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1

The sum to

n

terms of the series

{2^2} + {4^2} + {6^2} + ...

View answer and explanation

Correct option: B

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

Option B (

\frac{{2n(n + 1)(2n + 1)}}{3}

) is correct because the sum of the series

{2^2} + {4^2} + {6^2} + ...

n

terms is equal to

4

times the sum of the squares of the first

n

natural numbers.

Practice All Chapter MCQs →

A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}

{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}