Let Sn=∑r=1n tr=16n2n2+9n+13, then ∑r=1n tr equals

tr=Sr−Sr−1=26r3−(r−1)3+96r2−(r−1)2+136=(r+1)2Thus, ∑r=1n tr=∑r=1n (r+1)=n2(n+3)

Let Sn=∑r=1n tr=16n2n2+9n+13, then ∑r=1n tr equals

tr=Sr−Sr−1=26r3−(r−1)3+96r2−(r−1)2+136=(r+1)2Thus, ∑r=1n tr=∑r=1n (r+1)=n2(n+3)

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Let $S_{n} = \sum_{r = 1}^{n} t_{r} = \frac{1}{6} n (2 n^{2} + 9 n + 13)$ , then $\sum_{r = 1}^{n} \sqrt{t_{r}}$ equals

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

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

$(n + 1)^{2}$

$n^{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} t_{r} = S_{r} - S_{r - 1} \\ = \frac{2}{6} [r^{3} - (r - 1)^{3}] + \frac{9}{6} [r^{2} - (r - 1)^{2}] + \frac{13}{6} \\ = (r + 1)^{2} \end{array}$

Thus, $\sum_{r = 1}^{n} \sqrt{t_{r}} = \sum_{r = 1}^{n} (r + 1) = \frac{n}{2} (n + 3)$

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Let Sn=∑r=1n tr=16n2n2+9n+13, then ∑r=1n tr equals

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Let $S_{n} = \sum_{r = 1}^{n} t_{r} = \frac{1}{6} n (2 n^{2} + 9 n + 13)$ , then $\sum_{r = 1}^{n} \sqrt{t_{r}}$ equals