Animated Solution for Mathematics - Functions: For every positive integer $n$, prove that $\sqrt{(4n+1)}<\sqrt{n}+\sqrt{n+1}<\sqrt{4n+2}$. Hence or otherwise, prove that $[\sqrt{n}+\sqrt{n+1}]=[\sqrt{4n+1}]$, where $[x]$ denotes the greatest integer not exceeding $x$.

Welcome, future engineer. Today, we are going to tackle a problem that might look intimidating at first glance, but it is actually a beautiful exercise in precision and logical structure.

We are dealing with the expression $S=\sqrt{n}+\sqrt{n+1}$. Our goal is to prove that this sum is trapped between $\sqrt{4n+1}$ and $\sqrt{4n+2}$, and then use that to find its greatest integer value.

When you see a sum of square roots, your first instinct might be to reach for a calculator or try to estimate values. But in JEE Advanced, we do not estimate; we prove.

The most natural way to handle the sum of square roots is to square the entire expression. Let us define $S=\sqrt{n}+\sqrt{n+1}$. If we square both sides, we use the algebraic identity $(a+b{)}^2=a^2+b^2+2ab$.

Applying this to our expression, we get:

This simplifies beautifully to:

Now, look at this expression. The $2n+1$ part is fixed. The complexity lies entirely within the term $2\sqrt{n^2+n}$. To bound $S^2$, we must bound this inner radical.

This is where the magic happens. We need to find a range for $n^2+n$. We know that $n^2<n^2+n$.

To find an upper bound, we choose $(n+\frac{1}{2}{)}^2$. Expanding this, we get:

Notice the brilliance here! Since $\frac{1}{4}$ is positive, it is strictly true that:

Which means:

Taking the square root across this inequality, we find that $\sqrt{n^2+n}$ is strictly between $n$ and $n+\frac{1}{2}$. We have successfully constrained our radical.

Now, let us substitute these bounds back into our expression for $S^2$. We had $S^2=2n+1+2\sqrt{n^2+n}$.

For the lower bound:

For the upper bound:

So, we have established that $4n+1<S^2<4n+2$. Taking the square root of all parts, we arrive at the first part of our proof:

Finally, we address the greatest integer function, denoted by $[x]$. We want to find $[S]=[\sqrt{n}+\sqrt{n+1}]$.

Think about the numbers $4n+1$ and $4n+2$. They are consecutive integers. There is no perfect square between them.

Therefore, the square root of any number between $4n+1$ and $4n+2$ must have the same integer part. Since $S$ is trapped in this interval, its integer part must be the same as the integer part of $\sqrt{4n+1}$.

Thus, we conclude:

This is the elegance of mathematics. We did not need complex calculus or heavy machinery; we simply used the structure of the numbers themselves to squeeze the truth out of the expression. Keep this mindset—always look for the simplest, most elegant path—and you will conquer any problem JEE throws at you.