If an = -24n^2 - 16n + 15, then a1 + a2 + + a5 is equal to: — Solution | JEE Mathematics

MCQMediumJEE 2023Sum of Series

If $a_n = \frac{-2}{4n^2 - 16n + 15}, \quad then \quad a_1 + a_2 + \dots + a_5$ is equal to:

Answer

Correct answer:B

Standard Method

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

Find: $a_1 + a_2 + \cdots + a_5$

From the solution, the denominator is rewritten as

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

So,

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

Now use partial fractions:

\frac{-2}{(2n-3)(2n-5)} = \frac{1}{2n-3} - \frac{1}{2n-5}

Therefore,

\sum_{n=1}^{25} a_n = \sum_{n=1}^{25} \left(\frac{1}{2n-3} - \frac{1}{2n-5}\right)

This is a telescoping series, so most terms cancel and we get

\frac{1}{47} - \frac{1}{-3} = \frac{1}{47} + \frac{1}{3} = \frac{50}{141}

Therefore, the correct option is B according to the solution, even though the option text values do not align with this conclusion.

Factoring $4n^2 - 16n + 15$ incorrectly. This breaks the telescoping structure. First write it as $4n^2 - 6n - 10n + 15$ and then factor to get $(2n-3)(2n-5)$ .
Using the upper limit as $5$ from the question text when the solution works with $\sum_{n=1}^{25} a_n$ . This creates a mismatch between question and solution. Always check whether the extracted solution corresponds exactly to the displayed question.
Reversing the partial fractions. If $\frac{-2}{(2n-3)(2n-5)}$ is decomposed with the wrong sign, the telescoping cancellation gives the wrong result. Verify by recombining the fractions.

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