Question

Solve the question above with proper steps, correct opt is (b). ​

Answer 1

Answer:

(b)

Step-by-step explanation:

The k-th term of the series is

$\displaystyle\frac1{\sqrt{3k-1}+\sqrt{3k+2}}\\\\\\=\frac{\sqrt{3k-1}-\sqrt{3k+2}}{\bigl(\sqrt{3k-1}+\sqrt{3k+2}\bigr)\bigl(\sqrt{3k-1}-\sqrt{3k+2}\bigr)}\\\\\\=\frac{\sqrt{3k-1}-\sqrt{3k+2}}{(3k-1)-(3k+2)}\\\\\\=\tfrac13\bigl(\sqrt{3k+2}-\sqrt{3k-1}\bigr)$

So the sum of the first n terms is

$\displaystyle\tfrac13\bigl((\sqrt5-\sqrt2)+(\sqrt8-\sqrt5)+(\sqrt{11}-\sqrt8)+\dots+(\sqrt{3n+2}-\sqrt{3n-1})\bigr)\\\\=\tfrac13\bigl(\sqrt{3n+2}-\sqrt2\bigr)$