$$S=\left(\dfrac{1}{3}+\dfrac{4}{7}\right)+\left(\dfrac{1}{3^2}+\dfrac{1}{3}\times\dfrac{4}{7}+\dfrac{4^2}{7^2}\right)+\left(\dfrac{1}{3^3}+\dfrac{1}{3^2}\times\dfrac{4}{7}+\dfrac{1}{3}\times\dfrac{4^2}{7^2}+\dfrac{4^3}{7^3}\right)+......\infty$$
Take $$\dfrac{1}{3}=a$$ and $$\dfrac{4}{7}=b$$.
$$S=\left(a+b\right)+\left(a^2+a\times b+b^2\right)+\left(a^3+a^2\times b+a\times b^2+b^3\right)+......\infty$$
Multiply both sides by (a-b).
$$S\left(a-b\right)=\left(a+b\right)\left(a-b\right)+\left(a^2+a\times b+b^2\right)\left(a-b\right)+\left(a^3+a^2\times b+a\times b^2+b^3\right)\left(a-b\right)+......\infty$$
$$S\left(a-b\right)=\left(a^2-b^2\right)+\left(a^3-b^3\right)+\left(a^4-b^4\right)+......\infty$$
$$S\left(a-b\right)=\left(a^2+a^3+a^4+......\infty\right)-\left(b^2+b^3+b^4+........\infty\right)$$
We can see that $$\left(a^2+a^3+a^4+......\infty\right)$$ and $$\left(b^2+b^3+b^4+......\infty\right)$$ are in an infinite G.P.
$$\left(a^2+a^3+a^4+......\infty\right)=\dfrac{a^2}{1-a}$$
$$\left(b^2+b^3+b^4+......\infty\right)=\dfrac{b^2}{1-b}$$
$$S\left(a-b\right)=\left(\dfrac{a^2}{1-a}\right)-\left(\dfrac{b^2}{1-b}\right)$$
Now, insert the value of a and b in the above equation.
$$S\left(\dfrac{1}{3}-\dfrac{4}{7}\right)=\left(\dfrac{\left(\dfrac{1}{3}\right)^2}{1-\dfrac{1}{3}}\right)-\left(\dfrac{\left(\dfrac{4}{7}\right)^2}{1-\dfrac{4}{7}}\right)$$
$$S\left(\dfrac{1}{3}-\dfrac{4}{7}\right)=\left(\dfrac{\dfrac{1}{9}}{\dfrac{2}{3}}\right)-\left(\dfrac{\dfrac{16}{49}}{\dfrac{3}{7}}\right)$$
$$S\left(\dfrac{1}{3}-\dfrac{4}{7}\right)=\left(\dfrac{1}{6}\right)-\left(\dfrac{16}{21}\right)$$
$$S\left(\dfrac{7-12}{21}\right)=\dfrac{7-32}{42}$$
$$S\times\dfrac{-5}{21}=\dfrac{-25}{42}$$
$$S=\dfrac{5}{2}$$
Hence, the sum of the given expression is $$\dfrac{5}{2}$$.
$$\therefore\ $$ The required answer is C.
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