Question

The sum of the numerical series $\frac{1}{\sqrt{3}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{11}}+\frac{1}{\sqrt{11}+\sqrt{15}}+....$ upto $n$ terms, is

• A. $\frac{\sqrt{3+4n}-\sqrt{3}}{4}$
• B. $\frac{4}{\sqrt{3+4n}+\sqrt{3}}$
• C. Less than $n$
• D. Greater then $\frac{\sqrt{n}}{2}$

Answer 1

Answer: (A), (C)

$\frac{\sqrt{3+4n}-\sqrt{3}}{4}$
Less than $n$

Rationalizing each term:

$\frac{\sqrt{7}-\sqrt{3}}{7-3}+\frac{\sqrt{11}-\sqrt{7}}{11-7}+...+\frac{\sqrt{4n+3}-\sqrt{4n-1}}{(4n+3)-(4n-1)}=\frac{1}{4}.(\sqrt{7}-\sqrt{3}+\sqrt{11}-\sqrt{7}+...+\sqrt{4n+3}-\sqrt{4n-1})=\frac{1}{4}.(\sqrt{4n+3}-\sqrt{3})$

Rationalizing the above answer:

$\frac{(\sqrt{4n+3}-\sqrt{3})}{4}=\frac{(\sqrt{4n+3}-\sqrt{3})(\sqrt{4n+3}+\sqrt{3})}{4.(\sqrt{4n+3}-\sqrt{3})}=\frac{(4n+3-3)}{4.(\sqrt{4n+3}+\sqrt{3})}=\frac{n}{(\sqrt{4n+3}+\sqrt{3})}$

Since the result is $n$ divided by a positive quantity, it must be $less$ than $n$.

Again, if we assume the last option to be true, we have:

$\frac{n}{(\sqrt{4n+3}+\sqrt{3})}>\frac{\sqrt{n}}{2}=>\frac{\sqrt{n}}{(\sqrt{4n+3}+\sqrt{3})}>\frac{1}{2}=>\frac{n}{4n+3+3+2\sqrt{12n+9}}>\frac{1}{4}=>4n>4n+3+3+2\sqrt{12n+9}=>6+2\sqrt{12n+9}<0$

But this is false. Hence, the last option is wrong.

Hence, (a) and (c) are correct.