Question

Simplify: $\frac{\sqrt{3} + \sqrt{27} + \sqrt{75} + \sqrt{147} +\sqrt{243} +\sqrt{363} }{3\sqrt{3} }$

Answer 1

Step-by-step explanation:

Given that

$\frac{ \sqrt{3} + \sqrt{27} + \sqrt{75} + \sqrt{147} + \sqrt{243} + \sqrt{363} }{3 \sqrt{3} }$

To simplify it

Solution

As given expression we can write as

$\frac{ \sqrt{3} + \sqrt{3 \times 9} + \sqrt{3 \times 25} + \sqrt{3 \times 49} + \sqrt{3 \times 81} + \sqrt{3 \times 121} }{3 \sqrt{3} }$

as 9,25,49,81 and 121 are square of 3,5,7,9 and 11 respectively

$\frac{ \sqrt{3} + 3 \sqrt{3} + 5 \sqrt{3} + 7 \sqrt{3} + 9 \sqrt{3} + 11 \sqrt{3} }{3 \sqrt{3} }$

now we can take common

$\frac{(1 + 3 + 5 + 7 + 9 + 11) \times \sqrt{3} }{3 \sqrt{3} }$

by addition we get

$\frac{36 \sqrt{3} }{3 \sqrt{3} }$

by division we get

$12$

hence this is the answer.