Question

Let $S = \sqrt{1 + \frac{1}{1^2} + \frac{1}{2^2}} + \sqrt{1 + \frac{1}{2^2} + \frac{1}{3^2}} + \sqrt{1 + \frac{1}{3^2} + \frac{1}{4^2}} + ... + \sqrt{1 + \frac{1}{2024^2} + \frac{1}{2025^2}}$

then $\frac{(2025)S - 2024}{2024}$ is

Answer 1

2025

Answer 2

The problem asks us to evaluate an expression involving a sum $S$. The sum $S$ is given by:

$S = \sqrt{1 + \frac{1}{1^2} + \frac{1}{2^2}} + \sqrt{1 + \frac{1}{2^2} + \frac{1}{3^2}} + \sqrt{1 + \frac{1}{3^2} + \frac{1}{4^2}} + ... + \sqrt{1 + \frac{1}{2024^2} + \frac{1}{2025^2}}$

First, let's analyze the general term of the sum, $T_n = \sqrt{1 + \frac{1}{n^2} + \frac{1}{(n+1)^2}}$. We need to simplify the expression inside the square root: $1 + \frac{1}{n^2} + \frac{1}{(n+1)^2}$.

Combine the terms by finding a common denominator $n^2(n+1)^2$:

$1 + \frac{1}{n^2} + \frac{1}{(n+1)^2} = \frac{n^2(n+1)^2 + (n+1)^2 + n^2}{n^2(n+1)^2}$

Let's simplify the numerator:

$n^2(n+1)^2 + (n+1)^2 + n^2 = n^2(n^2+2n+1) + (n^2+2n+1) + n^2 = n^4 + 2n^3 + n^2 + n^2 + 2n + 1 + n^2 = n^4 + 2n^3 + 3n^2 + 2n + 1$

This expression is a perfect square. It is $(n^2+n+1)^2$. Let's verify:

$(n^2+n+1)^2 = ((n^2+n)+1)^2 = (n^2+n)^2 + 2(n^2+n)(1) + 1^2 = (n^4+2n^3+n^2) + (2n^2+2n) + 1 = n^4+2n^3+3n^2+2n+1$

So, the numerator is indeed $(n^2+n+1)^2$.

Therefore, the expression inside the square root becomes:

$1 + \frac{1}{n^2} + \frac{1}{(n+1)^2} = \frac{(n^2+n+1)^2}{n^2(n+1)^2} = \left(\frac{n^2+n+1}{n(n+1)}\right)^2$

Now, the general term $T_n$ is:

$T_n = \sqrt{\left(\frac{n^2+n+1}{n(n+1)}\right)^2}$

Since $n$ is a positive integer, $n(n+1)$ and $n^2+n+1$ are positive, so we can remove the square root:

$T_n = \frac{n^2+n+1}{n(n+1)}$

We can rewrite this expression by splitting the numerator:

$T_n = \frac{n(n+1)+1}{n(n+1)} = \frac{n(n+1)}{n(n+1)} + \frac{1}{n(n+1)} = 1 + \frac{1}{n(n+1)}$

Using partial fraction decomposition for $\frac{1}{n(n+1)}$:

$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$

So, the general term simplifies to:

$T_n = 1 + \left(\frac{1}{n} - \frac{1}{n+1}\right)$

Now, let's write out the sum $S$:

$S = \sum_{n=1}^{2024} T_n = \sum_{n=1}^{2024} \left(1 + \frac{1}{n} - \frac{1}{n+1}\right)$

This is a telescoping sum. Let's expand the terms:

$S = \left(1 + \frac{1}{1} - \frac{1}{2}\right) + \left(1 + \frac{1}{2} - \frac{1}{3}\right) + \left(1 + \frac{1}{3} - \frac{1}{4}\right) + ... + \left(1 + \frac{1}{2024} - \frac{1}{2025}\right)$

The '1' term appears 2024 times. The intermediate fractional terms cancel out:

$S = (1 \times 2024) + \left(1 - \frac{1}{2025}\right) = 2024 + 1 - \frac{1}{2025} = 2025 - \frac{1}{2025}$

Finally, we need to calculate the value of $\frac{(2025)S - 2024}{2024}$. First, calculate $(2025)S$:

$(2025)S = 2025 \left(2025 - \frac{1}{2025}\right) = 2025^2 - 2025 \times \frac{1}{2025} = 2025^2 - 1$

Now substitute this into the expression:

$\frac{(2025)S - 2024}{2024} = \frac{(2025^2 - 1) - 2024}{2024} = \frac{2025^2 - 1 - 2024}{2024} = \frac{2025^2 - 2025}{2024}$

Factor out 2025 from the numerator:

$\frac{2025(2025 - 1)}{2024} = \frac{2025 \times 2024}{2024}$

Cancel out 2024 from the numerator and denominator:

$= 2025$