Question

If the sum of the first ten terms of the series is (13/5) + (13/5) + (13/5) + 4 + (13/5) +…, is (16/5)m, then m is equal to:

Answer 1

Given the ambiguity of the question, and assuming it refers to a similar series pattern as in the related problem, m = 101.

Answer 2

The original question is ambiguous and likely contains a typo. The phrasing "If the sum of the first ten terms of the series is (13/5) + (13/5) + (13/5) + 4 + (13/5) +…, is (16/5)m" has two "is" clauses, leading to confusion and an undefined series.

Assuming the question intended to represent a series similar to the one in the related problem:

$\left(1 \frac{3}{5}\right)^{2} + \left(2 \frac{2}{5}\right)^{2} + \left(3 \frac{1}{5}\right)^{2} + 4^{2} + \left(4 \frac{4}{5}\right)^{2} + ....$

Which can be written as:

$\left(\frac{8}{5}\right)^{2} + \left(\frac{12}{5}\right)^{2} + \left(\frac{16}{5}\right)^{2} + \left(\frac{20}{5}\right)^{2} + \left(\frac{24}{5}\right)^{2} + ....$

The general term can be represented as $\left(\frac{4(n+1)}{5}\right)^2$.

The sum of the first 10 terms is:

$S_{10} = \sum_{n=1}^{10} \left(\frac{4(n+1)}{5}\right)^2 = \frac{16}{25} \sum_{n=1}^{10} (n+1)^2$

Let $k = n+1$, then:

$S_{10} = \frac{16}{25} \sum_{k=2}^{11} k^2$

Using the formula for the sum of squares: $\sum_{k=1}^{N} k^2 = \frac{N(N+1)(2N+1)}{6}$

$\sum_{k=2}^{11} k^2 = \left(\sum_{k=1}^{11} k^2\right) - 1^2 = \frac{11 \times 12 \times 23}{6} - 1 = 506 - 1 = 505$

Therefore:

$S_{10} = \frac{16}{25} \times 505$

Given $S_{10} = \frac{16}{5}m$:

$\frac{16}{25} \times 505 = \frac{16}{5}m$

Solving for $m$:

$m = \frac{505}{5} = 101$

Thus, $m = 101$.