Question

The rth term, $u_r$, of a series is given by $$u_r = (\frac{1}{3})^{3r-2} + (\frac{1}{3})^{3r-1}$$. Express $\sum_{r=1}^{n} u_r$ in the form $A(1-\frac{B}{27^n})$, where A and B are constants.

Name: the rth term ur of a series is given by ur frac133r 2 frac133r 1 express sumr1n ur in the form a1 fracb27n where a and b are constants 84518
Uploaded: 2020-04-08T12:30:51-08:00
Duration: 41 s
Channel: Tony Ni
Description: the rth term ur of a series is given by ur frac133r 2 frac133r 1 express sumr1n ur in the form a1 fracb27n where a and b are constants 84518

          The rth term, $u_r$, of a series is given by
$$u_r = (\frac{1}{3})^{3r-2} + (\frac{1}{3})^{3r-1}$$.
Express $\sum_{r=1}^{n} u_r$ in the form $A(1-\frac{B}{27^n})$, where A and B are constants.

$the rth term ur of a series is given by ur frac133r 2 frac133r 1 express sumr1n ur in the form a1 fracb27n where a and b are constants 84518$

Added by Sharon H.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Tony Ni

04/08/2020

Step 1

$$\sum_{r=1}^{n} u_r = \sum_{r=1}^{n} \left( (\frac{1}{3})^{3r-2} + (\frac{1}{3})^{3r-1} \right) = \sum_{r=1}^{n} (\frac{1}{3})^{3r-2} + \sum_{r=1}^{n} (\frac{1}{3})^{3r-1}$$ Let's consider the first sum: $$\sum_{r=1}^{n} (\frac{1}{3})^{3r-2} = Show more…

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The rth term, $u_r$, of a series is given by $$u_r = (\frac{1}{3})^{3r-2} + (\frac{1}{3})^{3r-1}$$. Express $\sum_{r=1}^{n} u_r$ in the form $A(1-\frac{B}{27^n})$, where A and B are constants.