Tính M = 1/2^3 + 2/3^3 + ... + 2022/2023^3

To find the sum of the series, we can use the formula for the sum of a finite geometric series:

\[S_n = \frac{a(1-r^n)}{1-r}\]

where:
- \(S_n\) is the sum of the first n terms
- \(a\) is the first term
- \(r\) is the common ratio
- \(n\) is the number of terms

In this case, the first term is \(1/2^3\), the common ratio is \(1/3\), and there are 2022 terms. So we have:

\[M = \frac{1/2^3(1 - (1/3)^{2022})}{1 - 1/3}\]

\[M = \frac{1/8(1 - 1/3)^{2022}}{2/3}\]

\[M = \frac{1/8(2/3)^{2022}}{2/3}\]

\[M = \frac{1}{8} \times \frac{(2/3)^{2022}}{2/3}\]

\[M = \frac{1}{8} \times \frac{2^{2022}/3^{2022}}{2/3}\]

\[M = \frac{1}{8} \times \frac{2^{2022}}{3^{2022}} \times \frac{3}{2}\]

\[M = \frac{3}{16} \times \left(\frac{2}{3}\right)^{2022}\]

Therefore, the sum of the series is \(\frac{3}{16} \times \left(\frac{2}{3}\right)^{2022}\).