frac{6^{2}+7^{2}+8^{2}+9^{2}+10^{2}}{sqrt{7+4 | vedclass

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Question

(B) First,simplify the denominator:$\sqrt{7+4 \sqrt{3}} = \sqrt{7+2 	imes 2 	imes \sqrt{3}} = \sqrt{4+3+2 	imes 2 	imes \sqrt{3}} = \sqrt{(2+\sqrt

Vedclass · Accepted Answer

$330$

Vedclass · Answer

(B) First,simplify the denominator:$\sqrt{7+4 \sqrt{3}} = \sqrt{7+2 	imes 2 	imes \sqrt{3}} = \sqrt{4+3+2 	imes 2 	imes \sqrt{3}} = \sqrt{(2+\sqrt{3})^{2}} = 2+\sqrt{3}$$\sqrt{4+2 \sqrt{3}} = \sqrt{3+1+2 	imes \sqrt{3} 	imes 1} = \sqrt{(\sqrt{3}+1)^{2}} = \sqrt{3}+1$Denominator $= (2+\sqrt{3}) - (\sqrt{3}+1) = 2+\sqrt{3}-\sqrt{3}-1 = 1$Next,calculate the numerator using the sum of squares formula $\sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}$:Sum of squares from $1$ to $10 = \frac{10(11)(21)}{6} = 385$Sum of squares from $1$ to $5 = \frac{5(6)(11)}{6} = 55$Numerator $= 6^{2}+7^{2}+8^{2}+9^{2}+10^{2} = 385 - 55 = 330$Result $= \frac{330}{1} = 330$

$\frac{6^{2}+7^{2}+8^{2}+9^{2}+10^{2}}{\sqrt{7+4 \sqrt{3}}-\sqrt{4+2 \sqrt{3}}}$ is equal to

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