The sum of first $ n $ terms of the given series $ 1^{2}+2 \cdot 2^{2}+3^{2} $ $ +2 \cdot 4^{2}+5^{2}+2 \cdot 6^{2}+\cdots $ is $ \frac{n(n+1)^{2}}{2} $, when $ n $ is even. When $ n $ is odd, the sum will be

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

The sum of first

n

terms of the given series

1^{2}+2 \cdot 2^{2}+3^{2}

+2 \cdot 4^{2}+5^{2}+2 \cdot 6^{2}+\cdots

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

, when

n

is even. When

n

is odd, the sum will be

View answer and explanation

Correct option: B

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

The correct answer is B (

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

). When n is odd, the sum of the series is derived by adding the nth term to the sum of the first n-1 terms, where n-1 is even.

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}