The sum of the series ${1^2} + 1 + {2^2} + 2 + {3^2} + 3 + \ldots + {n^2} + n$ is

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

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

$\frac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{4}$

Back to SEQUENCE OF SERIES

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

The sum of the series

{1^2} + 1 + {2^2} + 2 + {3^2} + 3 + \ldots + {n^2} + n

View answer and explanation

Correct option: C

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

Option C is correct because the sum of the series can be expressed as the sum of squares and the sum of natural numbers, which simplifies to the given expression.

More Questions from SEQUENCE OF SERIES

Practice All Chapter MCQs →

Given the sets

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

, then find the following
(i)

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

(ii)

A-(B \cap C)

(iii)

(B \cup C)-A

a,b,c

are in A.P.;

a,x,b

are in G.P. and

b,y,c

are in G.P., then

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

are in Suppose

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

is expressed as a polynomial in

y

, as

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

where

y=x-2

, then

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

equals If

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

and

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

are in

A P

, where

a, b

and c are in

G P

, then

a, b, c

are the lengths of sides of If the sum of three numbers in G.P. is 63 and the product of the first and the second term is

\frac{3}{4}

of the third term, then the numbers are If the lengths of sides of a right angled triangle are in A.P., then their ratio is