JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Analyzing +4 | -1

1^3=1,2^3=3+5,3^3=7+9+11

and

4^3=13+15+17+19

. then

n^3

as a similar series is

View answer and explanation

Correct option: C

\left(n^2-n+1\right)+\left(n^2-n+3\right)+\left(n^2-n+5\right)

+\ldots+\left(n^2+n-1\right)

Option C (

\left(n^2-n+1\right)+\left(n^2-n+3\right)+\left(n^2-n+5\right)+\ldots+\left(n^2+n-1\right)

) is correct because the series represents the sum of n consecutive odd numbers starting from n(n-1)+1.

More Questions from SEQUENCE OF SERIES

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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

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}