Animated Solution for Mathematics - Sequence and Series: For positive integers n, if $4a_n=(n^2+5n+6)$ and $S_n={\sum }_{k=1}^n(\frac{1}{a_k})$, then the value of $507S_{2025}$ is:

The beauty of the telescoping series is a powerful tool in your mathematical arsenal. We are given the expression $4a_n=n^2+5n+6$, and our objective is to determine the value of the sum $S_{2025}={\sum }_{k=1}^{2025}\frac{1}{a_k}$.

The first step in any sequence problem involving a quadratic is to factorize the expression. We examine $n^2+5n+6$, which can be factored by splitting the middle term:

Consequently, our expression for $a_n$ becomes:

Since we require the reciprocal for our summation, we invert the expression:

We now have a product of two consecutive integers in the denominator, which is the classic signature of a telescoping series. We can express the fraction as a difference by noting that $(k+3)-(k+2)=1$:

Multiplying by the constant $4$, we obtain the general term for our summation:

We now set up the sum $S_n=4{\sum }_{k=1}^n\left(\frac{1}{k+2}-\frac{1}{k+3}\right)$. Expanding this sum reveals a chain reaction of cancellations:

As the intermediate terms cancel out, we are left only with the first and the last terms:

We are now ready to calculate the value for $n=2025$. Substituting this into our formula, we get:

To subtract these, we find a common denominator. Since $2028=3\times 676$, we rewrite $\frac{1}{3}$ as $\frac{676}{2028}$:

Dividing $2028$ by $4$ yields $507$, simplifying the expression to $S_{2025}=\frac{675}{507}$. The final result, $507S_{2025}$, is therefore: