Infinite Arithmetic-Geometric Progression

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

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

$$1+\dfrac{1}{7}+\dfrac{1}{7^2}+\cdots+\infty$$ 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. 

Write the given infinite series as

$$S = 1 + \frac{2}{3} + \frac{6}{3^{2}} + \frac{10}{3^{3}} + \frac{14}{3^{4}} + \ldots$$

From the numerators we observe
$$1,\,2,\,6,\,10,\,14,\ldots$$
After the first two terms the numbers increase by $$4$$ each time. Hence for $$n \ge 2$$ the numerator can be written as $$4n-2$$, where $$n$$ is the power of $$3$$ in the denominator.

Therefore split the series:

$$S = 1 + \frac{2}{3} + \sum_{n=2}^{\infty} \frac{4n-2}{3^{n}}$$

Separate the two sums inside the sigma:

$$\sum_{n=2}^{\infty} \frac{4n-2}{3^{n}} = 4\sum_{n=2}^{\infty} \frac{n}{3^{n}} \;-\; 2\sum_{n=2}^{\infty} \frac{1}{3^{n}}$$

Let $$x = \frac13$$. We need the standard power-series results (valid for $$|x|\lt1$$):
$$\sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x},$$
$$\sum_{n=0}^{\infty} n x^{n} = \frac{x}{(1-x)^{2}}.$$

1. Geometric tail beginning with $$n=2$$: $$\sum_{n=2}^{\infty} x^{n} = \frac{x^{2}}{1-x}.$$ Putting $$x=\tfrac13$$ gives $$\sum_{n=2}^{\infty} \frac{1}{3^{n}} = \frac{(1/3)^{2}}{1-\tfrac13} = \frac{1/9}{2/3} = \frac16.$$

2. Arithmetico-geometric tail beginning with $$n=2$$: Start with the complete sum and subtract the $$n=1$$ term (the $$n=0$$ term is zero):
$$\sum_{n=2}^{\infty} n x^{n} = \frac{x}{(1-x)^{2}} - x.$$ Putting $$x=\tfrac13$$ gives $$\sum_{n=2}^{\infty} \frac{n}{3^{n}} = \frac{\tfrac13}{(2/3)^{2}} - \frac13 = \frac{1/3}{4/9} - \frac13 = \frac{3}{4} - \frac13 = \frac{5}{12}.$$

Now assemble the two parts:

$$\sum_{n=2}^{\infty} \frac{4n-2}{3^{n}} = 4\left(\frac{5}{12}\right) - 2\left(\frac16\right) = \frac{20}{12} - \frac{2}{6} = \frac{5}{3} - \frac13 = \frac{4}{3}.$$

Add the first two terms of the original series:

$$S = 1 + \frac{2}{3} + \frac{4}{3} = 1 + 2 = 3.$$

Therefore, the sum to infinity of the series is $$3$$.

Option B which is: $$3$$

Using the standard result,

$$S_n=2-\frac{n+2}{2^n}$$

Therefore,

$$2-S_n=\frac{n+2}{2^n}$$

Hence,

$$\sum_{n=1}^{\infty}(2-S_n) = \sum_{n=1}^{\infty}\frac{n+2}{2^n}$$

$$= \sum_{n=1}^{\infty}\frac{n}{2^n} + 2\sum_{n=1}^{\infty}\frac1{2^n}$$

Now,

$$\sum_{n=1}^{\infty}\frac{n}{2^n}=2$$

and

$$\sum_{n=1}^{\infty}\frac1{2^n}=1$$

Therefore,

$$\sum_{n=1}^{\infty}(2-S_n)=2+2$$

$$=4$$

Thus,

$$\frac{\lambda}{\mu}=4=\frac41$$

Hence,

$$\lambda+\mu=4+1$$

$$=5$$