S = ∑_k=1^n1√(3k+4)+√(3k+1) ⇒√(3k+4)+√(3k+1)/√((3k+4)^2)) (√(3k+1))^2 (on rationalizing denominator) √(3k+4) √(3k+1)3k+4 (3k+1) #160; S = ...

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Global Maxima

∑k=1n1√3k+4+√...

Question

$n \sum k = 1 \frac{1}{\sqrt{3 k + 4} + \sqrt{3 k + 1}}$

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Solution

$S = n \sum k = 1 \frac{1}{\sqrt{3 k + 4} + \sqrt{3 k + 1}}$
$\Rightarrow \frac{\sqrt{3 k + 4} + \sqrt{3 k + 1}}{\sqrt{(3 k + 4)^{2}}) - (\sqrt{3 k + 1})^{2}}$ (on rationalizing denominator)
$\frac{\sqrt{3 k + 4} - \sqrt{3 k + 1}}{3 k + 4 - (3 k + 1)}$
$S = \frac{1}{3} [n \sum k = 1 (\sqrt{3 k + 4} - \sqrt{3 k + 1})]$
$S = \frac{1}{3} [(\sqrt{7} - \sqrt{4}) + (\sqrt{10} - \sqrt{7}) + . . . . \sqrt{3 n + 4} - \sqrt{3 n + 1}]$
$S = \frac{1}{3} (\sqrt{3 n + 4} - \sqrt{4})$

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