Evaluate the series 2 8 32 128 512 2048.

Mathematics

1 answer:

Vikki [24]3 years ago

5 0

Answer: The given series is a geometric series sum 2730.

Step-by-step explanation: We are given to evaluate the following series :

2, 8, 32, 128, 512, 2048.

Let, $a_n$ denote the n-th term of the given series.

Then, we see the following pattern in the consecutive terms of the given series:

$a_1= 2, \\\\a_2=8=2\times 4=4a_1, \\\\a_3=32=8\times 4=4a_2\\\\a_4=128=32\times4=4a_3,\\\\a_5=2048=128\times4=4a_4,\\\\\vdots~~~~~~\vdots~~~~~~\vdots$

Therefore, each term after the first one is the product of the preceding term and 4.

That is, the given series is a geometric series with first term 2 and common ratio 4.

Thus, the required sum of the given series is

$S_5\\\\=2+8+32+128+512+2048\\\\=\dfrac{a(r^6-1)}{r-1}\\\\\\=\dfrac{2(4^6-1)}{4-1}\\\\\\=\dfrac{2}{3}\times (4096-1)\\\\=2\times1365\\\\=2730.$

The required sum of the series is 2730.

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VashaNatasha [74]

Answer:

Inside the circle, a distance of $\sqrt{34}$ from the center of the circle.

Step-by-step explanation:

Center of circle is at (0,0)

One point of the circle is at (0, $\sqrt{38}$ )

Since the only known point is in the y-axis, the radius of the circle must be $\sqrt{38}$ .

$\sqrt{38}$ = 6.164

So the point (-5,-3) is still inside the circle as the distance from point (-5,-3) to the center of the circle is $\sqrt{(5)^2+(-3)^2}$ or $\sqrt{34}$ which is less than the radius of the circle.