For any integer $n \ge 1$, the sum $\mathop \sum \limits_{k = 1}^n k\left( {k + 2} \right)$ is equal to

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

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

$\frac{{n\left( {n + 1} \right)\left( {2n + 9} \right)}}{6}$

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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1

For any integer

n \ge 1

, the sum

\sum \limits_{k = 1}^n k\left( {k + 2} \right)

is equal to

View answer and explanation

Correct option: C

\frac{{n\left( {n + 1} \right)\left( {2n + 7} \right)}}{6}

Option C (n(n+1)(2n+7)/6) is correct because it is the simplified form of the summation after applying the formulas for the sum of first n natural numbers and the sum of squares.

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