Arithmetic-Geometric Progression

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,

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

$$=5$$