**Brief Solution:**

Step 1 – Let’s find the sum = 2π(1) + 2π(4/5) + 2π(16/25) … = 2π(1 + 4/5 + 16/25 + ……..)

Step 2 – If we look at the sum inside the brackets, it is nothing but a GP, whose sum we know.

Step 3 – S = 2π(1 + 4/5 + 16/25 ……) = 2π(5) = 10π

**Detailed Solution:**

**Step1:**

If we look at the radii of the circles, then the following series will become apparent:

1, 4/5, 16/25, 64/125, …

This series can be rewritten as:

1, (4/5), (4/5)x(4/5), (4/5)x(4/5)x(4/5), …

We can see that this is a geometric series where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio.

In our case, the first term is 1 and the common ratio is 4/5.

Also, from the knowledge of geometric series, we know that if the series is infinite and the absolute value of the common ratio is less than 1, then the sum of the series is given by:

S = a/(1-r)

where,

a is the first term of the sequence

r is the common ratio

Therefore,

S = 1/(1 – 4/5)

S = 1/(1/5)

S = 5

**Step 2:**

We know that the circumference of a circle is 2πR, where R is the radius of the circle.

**Step 3:**

Now, we calculate the sum of the circumferences of the infinite number of concentric circles:

S’ = 2π(1) + 2π(4/5) + 2π(4/5)(4/5) + 2π(4/5)(4/5)(4/5)……

Taking 2π as common:

S’ = 2π {1 + (4/5) + (4/5)(4/5) + (4/5)(4/5)(4/5)…..}

From Step 1, we know that:

The sum of 1 + (4/5) + (4/5)(4/5) + (4/5)(4/5)(4/5) + …equals 5.

Therefore, the sum of the circumferences of the infinite number of concentric circles is:

2π(5) = 10π