JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Analyzing +4 | -1

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

, then

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

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because after finding the general term

t_r

and summing the square root of the terms, the result simplifies to

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

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