If ${S_n} = \mathop \sum \limits_{r = 1}^n {a_r} = \frac{1}{6}n\left( {2{n^2} + 9n + 13} \right),$ then $\mathop \sum \limits_{r = 1}^n \sqrt {{a_r}} $ equals

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

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

$\frac{{n\left( {n + 5} \right)}}{2}$

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Analyzing +4 | -1

{S_n} = \sum \limits_{r = 1}^n {a_r} = \frac{1}{6}n\left( {2{n^2} + 9n + 13} \right),

then

\sum \limits_{r = 1}^n \sqrt {{a_r}}

equals

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is derived by finding the general term

{a_r}

and then summing the square root of the general term.

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}