The sum $\displaystyle\sum_{n=1}^{21} \frac{3}{(4 n-1)(4 n+3)}$ is equal to

Question

Let \[ S=\frac{1}{2!5!}+\frac{1}{3!2!3!}+\frac{1}{5!2!1!}+\cdots \text{ up to 13 terms}. \] If $13S=\dfrac{2^k}{n!}$, $k\in\mathbb{N}$, then $n+k$ is equal to

Answer 1

To solve the given problem, we need to evaluate the sum:

$\sum_{n=1}^{21} \frac{3}{(4n-1)(4n+3)}$

First, we simplify the given expression by partial fraction decomposition:

We write:

$\frac{3}{(4n-1)(4n+3)} = \frac{A}{4n-1} + \frac{B}{4n+3}$

where A and B are constants to be determined. Clearing the denominators gives:

$3 = A(4n+3) + B(4n-1)$

Expanding and rearranging the equation:

$3 = (4A + 4B)n + (3A - B)$

For the equality to hold for all n, the coefficients of n and the constant terms must separately be equal:

$4A + 4B = 0$
$3A - B = 3$

From (1), we have:

$A = -B$

Substituting into (2):

$3(-B) - B = 3$

Simplifying gives:

$-4B = 3 \implies B = -\frac{3}{4}$

Thus, $A = \frac{3}{4}$.

Substitute A and B back into the partial fractions:

$\frac{3}{(4n-1)(4n+3)} = \frac{\frac{3}{4}}{4n-1} - \frac{\frac{3}{4}}{4n+3}$

This leads to a telescoping series:

$\sum_{n=1}^{21} \left( \frac{3}{4(4n-1)} - \frac{3}{4(4n+3)} \right)$

Terms will cancel out except the first negative and the last positive terms:

$\frac{3}{4} \left( \frac{1}{3} - \frac{1}{87} \right)$

Calculate the simplified result:

$\frac{3}{4} \times \frac{86}{261}$

Simplifying further:

$(\frac{3 \times 86}{4 \times 261} = \frac{258}{1044} = \frac{7}{29}$

Therefore, the sum is:

$\frac{7}{29}$

Thus, the correct answer is:

$\frac{7}{29}$

The sum $\displaystyle\sum_{n=1}^{21} \frac{3}{(4 n-1)(4 n+3)}$ is equal to

The Correct Option is B

Solution and Explanation

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