Mathematics: For n prove by thank you

For n prove by thank you - id7194208, HoneySoft

Ответ:

libi052207

30.01.2022

Mathematics

$3 + 3^2 + 3^3 ...+ 3^n = \frac{3(3^n - 1)}{2}$

Step-by-step explanation:

Given

3 + 3^2 + 3^3 ...+ 3^n

Required

Show that the sum of the series is $\frac{3(3^{n}-1)}{2}$

The above shows the sum of a geometric series and this will be calculated as shown below;

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

Where

a = First term;

a = 3

r = common ratio

$r = \frac{3^2}{3} = \frac{3^3}{3^2}$

r = 3

Substitute 3 for r and 3 for a in the formula above;

$S_n = \frac{a(r^n - 1)}{r - 1}$ becomes

$S_n = \frac{3(3^n - 1)}{3 - 1}$

$S_n = \frac{3(3^n - 1)}{2}$

Hence;

$3 + 3^2 + 3^3 ...+ 3^n = \frac{3(3^n - 1)}{2}$

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Ответ:

lucywood2024

30.11.2020

Mathematics

Sample Response: If the data is collected in a biased manner, the graph could be skewed. Also, if the number of observations is too small, the graph can be skewed. To ensure the data is not skewed, collect a large representative unbiased sample

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For n prove by $3 + 3 {}^{2} + 3 {}^{3} ... + 3 {}^{n} = \frac{3(3 {}^{n} - 1) }{2}$
thank you

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