JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

The sum of first

n

terms of the series

1^2+2\left(2^2\right)+3^2+2\left(4^2\right)+5^2+2\left(6^2\right)+\ldots

n(n+1)^2 / 2

when

n

is even. When

n

is odd, the sum is

View answer and explanation

Correct option: B

n^2(n+1) / 2

Option C (

\frac{{1}}{{3}}

) is correct. By setting up the equations based on the given information and solving for the common ratio, we find r = 1/3.

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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

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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}