Infinite Geometric Progression

$$S=\left(\dfrac{1}{3}+\dfrac{4}{7}\right)+\left(\dfrac{1}{3^2}+\dfrac{1}{3}\times\dfrac{4}{7}+\dfrac{4^2}{7^2}\right)+\left(\dfrac{1}{3^3}+\dfrac{1}{3^2}\times\dfrac{4}{7}+\dfrac{1}{3}\times\dfrac{4^2}{7^2}+\dfrac{4^3}{7^3}\right)+......\infty$$

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. 

$$7=5+\dfrac{1}{7}(5+\alpha)+\dfrac{1}{7^2}(5+2\alpha)+\dfrac{1}{7^3}(5+3\alpha)+\cdots+\infty$$

$$7=5\left(1+\dfrac{1}{7}+\dfrac{1}{7^2}+\cdots+\infty\right)+\alpha\ \left(\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\ \right)$$

$$1+\dfrac{1}{7}+\dfrac{1}{7^2}+\cdots+\infty$$ is an infinite G.P with the common ratio of $$\dfrac{1}{7}$$.

Sum of the above G.P = $$\dfrac{1}{1-\dfrac{1}{7}}=\dfrac{7}{6}$$

$$7=\left(5\times\dfrac{7}{6}\right)+\alpha\ \left(\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\ \right)$$

$$7-\dfrac{35}{6}=\alpha\ \left(\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\ \right)$$

$$\dfrac{7}{6}=\alpha\ \left(\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\ \right)$$

Now, $$S=\left(\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\right)$$ in the form of AGP where $$\left(1,\ 2,\ 3,\ \cdots,\ \infty\right)$$ in A.P and $$\left(\dfrac{1}{7},\ \dfrac{1}{7^2},\ \dfrac{1}{7^3},\ \cdots,\ \infty\right)$$ are in G.P. 

$$S=\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\ $$                                        $$\longrightarrow\ i$$

Multiply both sides by $$\dfrac{1}{7}$$. 

$$\dfrac{S}{7}=\dfrac{1}{7^2}+\dfrac{2}{7^3}+\dfrac{3}{7^4}+\cdots+\infty\ $$                       $$\longrightarrow\ ii$$

Subtract equation $$ii$$ from equation $$i$$, 

$$\dfrac{6S}{7}=\dfrac{1}{7}+\left(\dfrac{2}{7^2}-\dfrac{1}{7^2}\right)+\left(\dfrac{3}{7^3}-\dfrac{2}{7^3}\right)+\left(\dfrac{4}{7^4}-\dfrac{3}{7^4}\right)+\cdots+\infty$$

$$\dfrac{6S}{7}=\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+\dfrac{1}{7^4}+\cdots+\infty$$

$$\dfrac{6S}{7}=\dfrac{\dfrac{1}{7}}{\left(1-\dfrac{1}{7}\right)}$$

$$\dfrac{6S}{7}=\dfrac{1}{6}$$

$$S=\dfrac{7}{36}$$

Hence, $$\dfrac{7}{6}=\alpha\ \times\dfrac{7}{36}$$

$$\alpha\ =6$$

Hence, the value of $$\alpha$$ is 6. 
$$\therefore\ $$ The required answer is B.