Sequence And Series Question 533
Question: The sum of the series $ \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+…+\frac{1}{\sqrt{n^{2}-1}+\sqrt{n^{2}}} $ equals
[AMU 2002]
Options:
A) $ \frac{(2n+1)}{\sqrt{n}} $
B) $ \frac{\sqrt{n}+1}{\sqrt{n}+\sqrt{n-1}} $
C) $ \frac{(n+\sqrt{n^{2}-1})}{2\sqrt{n}} $
D) $ n-1 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+….+\frac{1}{\sqrt{n^{2}}+\sqrt{n^{2}-1}} $ Rationalization of $ D^{r} $
$ \therefore S=(\sqrt{2}-\sqrt{1})+( \sqrt{3}-\sqrt{2} )+…+( \sqrt{n^{2}}-\sqrt{n^{2}-1} ) $ S = n - 1.
BETA
