Question

найти сумму бесконечной геометрической прогрессии, если 4√2; 4; 2√2;... найти S-?

Answer 1

$b_1=4\sqrt{2}; b_2=4;\\q=\frac{b_2}{b_1}=\frac{4}{4\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2};\\|q|<1;$

следовательно данная геометрическая прогрессия убывающая, по формуле суммы членов бесконечной убывающей геометричесской прогресиии

$S=\frac{b_1}{1-q}=\frac{4\sqrt{2}}{1-\frac{1}{\sqrt{2}}}=\frac{4\sqrt{2}*\sqrt{2}}{\sqrt{2}-1}=\frac{8}{\sqrt{2}-1}=\frac{8*(\sqrt{2}+1)}{2-1}=8*(\sqrt{2}+1)$

Answer 2

Calculation of the Sum of an Infinite Geometric Progression

To find the sum of an infinite geometric progression, we need to know the first term and the common ratio. In this case, the first term is 4√2, and the common ratio is 0.5.

Let's calculate the sum step by step:

Step 1:

The first term of the geometric progression is 4√2. The common ratio is 0.5.

Step 2:

Using the formula for the sum of an infinite geometric progression, we have:

S = a / (1 - r)

where S is the sum, a is the first term, and r is the common ratio.

Substituting the values, we get:

S = 4√2 / (1 - 0.5)

Step 3:

Simplifying the expression:

S = 4√2 / 0.5

Step 4:

Dividing 4√2 by 0.5:

S = 8√2

Therefore, the sum of the given infinite geometric progression is 8√2.

Please let me know if there's anything else I can help you with!