If $$a_n = \frac{-2}{4n^2 - 16n + 15}$$, then $$a_1 + a_2 + \ldots + a_{25}$$ is equal to:

To find the sum of the first 25 terms of the given series, we first rewrite the general term $$a_n$$ using partial fractions to create a telescoping series.

The general term is given as:

$$a_n = \frac{-2}{4n^2 - 16n + 15}$$

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Step 1: Factorise the denominator

The quadratic expression in the denominator can be factored by splitting the middle term:

$$4n^2 - 16n + 15 = 4n^2 - 6n - 10n + 15 = 2n(2n - 3) - 5(2n - 3) = (2n - 3)(2n - 5)$$

Substituting this back into the expression for $$a_n$$:

$$a_n = \frac{-2}{(2n - 3)(2n - 5)}$$

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Step 2: Express $$a_n$$ in partial fractions

We can rewrite the numerator -2 as the difference between the two factors in the denominator:

$$-2 = (2n - 5) - (2n - 3)$$

Substituting this into $$a_n$$:

$$a_n = \frac{(2n - 5) - (2n - 3)}{(2n - 3)(2n - 5)} = \frac{2n - 5}{(2n - 3)(2n - 5)} - \frac{2n - 3}{(2n - 3)(2n - 5)}$$

$$a_n = \frac{1}{2n - 3} - \frac{1}{2n - 5}$$

This gives the telescoping structure $$a_n = V_n - V_{n-1}$$ where $$V_n = \frac{1}{2n - 3}$$.

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Step 3: Expand the sum up to 25 terms

Now, we evaluate the individual terms of the series from n = 1 to n = 25:

$$a_1 = \frac{1}{-1} - \frac{1}{-3} = -1 + \frac{1}{3}$$

$$a_2 = \frac{1}{1} - \frac{1}{-1} = 1 + 1$$

$$a_3 = \frac{1}{3} - \frac{1}{1} = \frac{1}{3} - 1$$

$$a_4 = \frac{1}{5} - \frac{1}{3}$$

$$\vdots$$

$$a_{24} = \frac{1}{45} - \frac{1}{43}$$

$$a_{25} = \frac{1}{47} - \frac{1}{45}$$

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Step 4: Cancel the intermediate terms

Summing all the terms together:

$$\sum_{n=1}^{25} a_n = \left(-1 + \frac{1}{3}\right) + (1 + 1) + \left(\frac{1}{3} - 1\right) + \left(\frac{1}{5} - \frac{1}{3}\right) + \dots + \left(\frac{1}{47} - \frac{1}{45}\right)$$

Grouping the positive and negative terms reveals the cancellation pattern:

$$\sum_{n=1}^{25} a_n = \left(\frac{1}{3} + 1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{47}\right) - \left(1 - \frac{1}{3} - 1 + 1 + \frac{1}{3} + \dots + \frac{1}{45}\right)$$

Most of the intermediate fractions cancel out, leaving behind only the following terms:

$$\sum_{n=1}^{25} a_n = \frac{1}{3} + \frac{1}{47}$$

$$\sum_{n=1}^{25} a_n = \frac{47 + 3}{3 \times 47} = \frac{50}{141}$$

Therefore, the value of $$a_1 + a_2 + \ldots + a_{25}$$ is equal to $$\frac{50}{141}$$.