First slide

Series of natural numbers

Question

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

Moderate

A

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

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

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

$\frac{1}{2} n (n + 5)$

Solution

We have $t_{n} = S_{n} - S_{n - 1} \forall n \geq 2$
$\begin{array}{l} ∴ t_{n} = \frac{1}{6} n (2 n^{2} + 9 n + 13) - \frac{1}{6} (n - 1) \{(2 (n - 1)^{2} \\ = \frac{1}{6} [2 (n^{3} - (n - 1)^{3}) + 9 (n^{2} - (n - 1)^{2}) \\ = \frac{1}{6} [6 n^{2} - 6 n + 2 + 9 (2 n - 1) + 13] \\ = \frac{1}{6} (6 n^{2} + 12 n + 6) \\ = (n + 1)^{2} \end{array}$
Also, $t_{1} = S_{1} = 4 = (1 + 1)^{2}$
$∴ \sum_{r = 1}^{n} \sqrt{t_{r}} = \sum_{r = 1}^{n} (r + 1) = \frac{1}{2} (n + 1) (n + 2) - 1 = \frac{1}{2} n (n + 3)$

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