The value of sum

s = (1/3 + 4/7) + [(1/3)² + (4/7)² + (1/3)(4/7)]

+ [(1/3)³ + (1/3)²(4/7) + (1/3)(4/7)² + (4/7)³] + ...

then s is equal to

To find the value of the given infinite sum, let us first identify the pattern of the terms.

Let $a=\frac{1}{3}$ and $b=\frac{4}{7}$.

The given series $s$ can be rewritten as:
$s=(a+b)+(a^2+ab+b^2)+(a^3+a^{2}b+a{b}^2+b^3)+...$

We know the algebraic identity:
$a^{n+1}-b^{n+1}=(a-b)(a^n+a^{n-1}b+...+a{b}^{n-1}+b^n)$

Since $a\ne b$, we can write the terms of the series as:
$a+b=\frac{a^2-b^2}{a-b}$
$a^2+ab+b^2=\frac{a^3-b^3}{a-b}$
$a^3+a^{2}b+a{b}^2+b^3=\frac{a^4-b^4}{a-b}$

Substituting these back into the expression for $s$:
$s=\frac{a^2-b^2}{a-b}+\frac{a^3-b^3}{a-b}+\frac{a^4-b^4}{a-b}+...$

We can factor out $\frac{1}{a-b}$ and group the terms containing $a$ and $b$ separately:
$s=\frac{1}{a-b}\left[(a^2+a^3+a^4+...)-(b^2+b^3+b^4+...)\right]$

Since $\left|a\right|<1$ and $\left|b\right|<1$, both bracketed series are infinite geometric progressions with common ratios $a$ and $b$ respectively. Using the sum formula for an infinite geometric series $S=\frac{\text{first term}}{1-\text{ratio}}$, we have:
$a^2+a^3+a^4+...=\frac{a^2}{1-a}$
$b^2+b^3+b^4+...=\frac{b^2}{1-b}$

Thus, the sum becomes:
$s=\frac{1}{a-b}\left[\frac{a^2}{1-a}-\frac{b^2}{1-b}\right]$

Now, let's substitute the values $a=\frac{1}{3}$ and $b=\frac{4}{7}$:
$a-b=\frac{1}{3}-\frac{4}{7}=\frac{7-12}{21}=-\frac{5}{21}$

Calculate the terms inside the bracket:
$\frac{a^2}{1-a}=\frac{{(1/3)}^2}{1-1/3}=\frac{1/9}{2/3}=\frac{1}{9}\times \frac{3}{2}=\frac{1}{6}$
$\frac{b^2}{1-b}=\frac{{(4/7)}^2}{1-4/7}=\frac{16/49}{3/7}=\frac{16}{49}\times \frac{7}{3}=\frac{16}{21}$

Now find the difference:
$\frac{a^2}{1-a}-\frac{b^2}{1-b}=\frac{1}{6}-\frac{16}{21}=\frac{7-32}{42}=-\frac{25}{42}$

Substitute these values back into the expression for $s$:
$s=\frac{1}{-5/21}\times \left(-\frac{25}{42}\right)=\left(-\frac{21}{5}\right)\times \left(-\frac{25}{42}\right)$

Simplifying the fraction multiplication:
$s=\frac{21}{42}\times \frac{25}{5}=\frac{1}{2}\times 5=\frac{5}{2}$

Therefore, the correct option is 5/2.