The $$ r $$ th term, $$ u_{r} $$, of an infinite series is given by $$ u_{r}=\left(\frac{1}{3}\right)^{2 r+1}+\left(\frac{1}{3}\right)^{2 r-1} $$. (a) Express $$ u_{r} $$, in the form $$ \frac{A}{3^{2 r+1}} $$, where $$ A $$ is a constant. (b) Find the sum of the first $$ n $$ terms of the series, and deduce the sum of the infinite series.

Question

UpStudy AI · Accepted Answer

To solve the problem, we start with the expression for the $$ r $$ th term of the series:

$$
u_{r} = \left(\frac{1}{3}\right)^{2r+1} + \left(\frac{1}{3}\right)^{2r-1}
$$

### Part (a)

We want to express $$ u_{r} $$ in the form $$ \frac{A}{3^{2r+1}} $$.

First, we can rewrite the second term:

$$
\left(\frac{1}{3}\right)^{2r-1} = \frac{1}{\left(\frac{1}{3}\right)^{2r+1}} \cdot \left(\frac{1}{3}\right)^{2} = \frac{1}{3^{2r+1}} \cdot 3^2 = \frac{9}{3^{2r+1}}
$$

Now substituting this back into the expression for $$ u_{r} $$:

$$
u_{r} = \left(\frac{1}{3}\right)^{2r+1} + \frac{9}{3^{2r+1}} = \frac{1}{3^{2r+1}} + \frac{9}{3^{2r+1}} = \frac{1 + 9}{3^{2r+1}} = \frac{10}{3^{2r+1}}
$$

Thus, we have:

$$
u_{r} = \frac{10}{3^{2r+1}}
$$

where $$ A = 10 $$.

### Part (b)

Next, we need to find the sum of the first $$ n $$ terms of the series, $$ S_n = u_1 + u_2 + \ldots + u_n $$.

Using the expression we found for $$ u_r $$:

$$
S_n = \sum_{r=1}^{n} u_r = \sum_{r=1}^{n} \frac{10}{3^{2r+1}}
$$

Factoring out the constant $$ 10 $$:

$$
S_n = 10 \sum_{r=1}^{n} \frac{1}{3^{2r+1}} = 10 \sum_{r=1}^{n} \frac{1}{3^{2r} \cdot 3} = \frac{10}{3} \sum_{r=1}^{n} \frac{1}{3^{2r}}
$$

The series $$ \sum_{r=1}^{n} \frac{1}{3^{2r}} $$ is a geometric series with the first term $$ a = \frac{1}{3^2} = \frac{1}{9} $$ and common ratio $$ r = \frac{1}{9} $$:

$$
\sum_{r=1}^{n} \frac{1}{3^{2r}} = \frac{\frac{1}{9} \left(1 - \left(\frac{1}{9}\right)^{n}\right)}{1 - \frac{1}{9}} = \frac{\frac{1}{9} \left(1 - \frac{1}{9^n}\right)}{\frac{8}{9}} = \frac{1}{8} \left(1 - \frac{1}{9^n}\right)
$$

Substituting this back into the expression for $$ S_n $$:

$$
S_n = \frac{10}{3} \cdot \frac{1}{8} \left(1 - \frac{1}{9^n}\right) = \frac{10}{24} \left(1 - \frac{1}{9^n}\right) = \frac{5}{12} \left(1 - \frac{1}{9^n}\right)
$$

Now, to find the sum of the infinite series, we take the limit as $$ n \to \infty $$:

$$
S = \lim_{n \to \infty} S_n = \frac{5}{12} \left(1 - 0\right) = \frac{5}{12}
$$

### Final Answers

(a) $$ u_{r} = \frac{10}{3^{2r+1}} $$

(b) The sum of the infinite series is $$ \frac{5}{12} $$.
(a) $$ u_{r} = \frac{10}{3^{2r+1}} $$ (b) The sum of the infinite series is $$ \frac{5}{12} $$.

The th term, , of an infinite series is given by .
(a) Express , in the form , where is a constant.
(b) Find the sum of the first terms of the series, and deduce the sum of
the infinite series.

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Bonus Knowledge

(a) Express in the form .

(b) Find the sum of the first terms of the series and deduce the sum of the infinite series.

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The th term, , of an infinite series is given by . (a) Express , in the form , where is a constant. (b) Find the sum of the first terms of the series, and deduce the sum of the infinite series.