All categories

3 years ago

Find S11 for 1 + 2 + 4 + 8 +

1 answer:

4 0

Evaluating the given sequence, it is evident that the next number is twice the number prior to it. Thus, the given is a geometric sequence with first term (a1) equal to 1 and common ratio of 2. The geometric series may be calculated by the equation,

Sn = a1 x (1 - r^n) / (1 - r)

where Sn is the sum of n terms in this case, n = 11.
Substituting the known values,

<span> Sn = 1 x (1 - 2^11) / (1 - 2) = 2047
</span>
Thus, S11 is 2047.

You might be interested in

Stephen biked 75% of 2 miles fred biked 2.5 miles how much farther did Fred bike than Stephen

Natalka [10]

The answer is:

2.5 - (.75 • 2)
= 2.5 - 1.5
= 1

7 0

2 years ago

Hector spent $41.75 for 2 DVDs that cost the same amount. The sales tax on his purchase was $3.15. Hector also used a coupon for

kykrilka [37]

Given:

Amount spent on 2 DVDs = $41.75

Sales tax = $3.15

Coupon = $1.00 off

To find:

The cost of each DVD.

Solution:

Let x be the cost of each DVD.

Then, cost of two DVDs = 2x

Amount spent of 2 DVDs = Cost of 2 DVDs + Sales tax - Coupon discount

$41.75=2x+3.15-1.00$

$41.75=2x+2.15$

$41.75-2.15=2x$

$39.6=2x$

Divide both sides by 2.

$\dfrac{39.6}{2}=x$

$19.8=x$

Therefore, the cost of each DVD is $19.8.

8 0

3 years ago

What is the formula for the following geometric sequence? -5, -10, -20, -40, ... an = -5 · 2n - 1 an = -5 · (-2)n - 1 an = -2 ·

vazorg [7]

$-5 * 2^{n-1}$

8 0

3 years ago

Li Juan solves the equation below by first squaring both sides of the equation.

emmasim [6.3K]

-3w + 13
Happy New Year

8 0

3 years ago

Read 2 more answers

Find the radius of convergence, r, of the series. ∞ xn 2n − 1 n = 1 r = 1 find the interval, i, of convergence of the series. (e

Bingel [31]

Assuming the series is

$\displaystyle\sum_{n\ge1}\frac{x^n}{2n-1}$

The series will converge if

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|$

We have

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|=|x|\lim_{n\to\infty}\frac{\frac1{2n+1}}{\frac1{2n-1}}=|x|-\lim_{n\to\infty}\frac{2n-1}{2n+1}=|x|$

So the series will certainly converge if $-1$ , but we also need to check the endpoints of the interval.

If $x=1$ , then the series is a scaled harmonic series, which we know diverges.

On the other hand, if $x=-1$ , by the alternating series test we can show that the series converges, since

$\left|\dfrac{(-1)^n}{2n-1}\right|=\dfrac1{2n-1}\to0$

and is strictly decreasing.

So, the interval of convergence for the series is $-1\le x$ .

6 0

3 years ago