Geometric Progression: Sum to Infinity

Find $$\sum_{k=1}^{\infty} c_k - (12a_6 + 8b_4)$$ where $$c_k = a_k + b_k$$ with given conditions.

$$c_2 = 4r_1 + 4r_2 = 5 \Rightarrow r_1 + r_2 = \frac{5}{4}$$

$$c_3 = 4r_1^2 + 4r_2^2 = \frac{13}{4} \Rightarrow r_1^2 + r_2^2 = \frac{13}{16}$$

$$(r_1 + r_2)^2 = r_1^2 + 2r_1 r_2 + r_2^2$$

$$\frac{25}{16} = \frac{13}{16} + 2r_1 r_2 \Rightarrow r_1 r_2 = \frac{3}{8}$$

$$r_1, r_2$$ are roots of $$t^2 - \frac{5}{4}t + \frac{3}{8} = 0$$, i.e., $$8t^2 - 10t + 3 = 0$$.

$$t = \frac{10 \pm 2}{16}$$, so $$r_1 = \frac{1}{2}$$, $$r_2 = \frac{3}{4}$$ (since $$r_1 < r_2$$).

$$\sum_{k=1}^{\infty} c_k = \frac{4}{1 - r_1} + \frac{4}{1 - r_2} = \frac{4}{1/2} + \frac{4}{1/4} = 8 + 16 = 24$$

$$a_6 = 4 \cdot \left(\frac{1}{2}\right)^5 = \frac{4}{32} = \frac{1}{8}$$

$$b_4 = 4 \cdot \left(\frac{3}{4}\right)^3 = 4 \cdot \frac{27}{64} = \frac{27}{16}$$

$$12a_6 + 8b_4 = 12 \cdot \frac{1}{8} + 8 \cdot \frac{27}{16} = \frac{3}{2} + \frac{27}{2} = 15$$

$$\sum c_k - (12a_6 + 8b_4) = 24 - 15 = 9$$

The answer is $$\boxed{9}$$.

The given infinite geometric series converge because $$|a| \lt 1,\; |b| \lt 1,\; |c| \lt 1$$.

For an infinite G.P. with first term $$1$$ and common ratio $$r$$ (where $$|r| \lt 1$$), the sum is $$\displaystyle\frac{1}{1-r}$$. Applying this:

$$x=\sum_{n=0}^{\infty}a^{n}=\frac{1}{1-a},\qquad y=\sum_{n=0}^{\infty}b^{n}=\frac{1}{1-b},\qquad z=\sum_{n=0}^{\infty}c^{n}=\frac{1}{1-c}.$$

To check whether $$x,y,z$$ form an H.P., look at their reciprocals. We obtain

$$\frac{1}{x}=1-a,\qquad \frac{1}{y}=1-b,\qquad \frac{1}{z}=1-c.$$

Given that $$a,b,c$$ are in arithmetic progression, there exists a common difference $$d$$ such that $$b-a=d,\quad c-b=d.$$

Then $$(1-b)-(1-a)=a-b=-d,\qquad (1-c)-(1-b)=b-c=-d,$$ so $$1-a,\;1-b,\;1-c$$ are also in arithmetic progression.

Thus $$\displaystyle\frac{1}{x},\;\frac{1}{y},\;\frac{1}{z}$$ form an A.P. Therefore, $$x,y,z$$ form a harmonic progression (H.P.).

Option D which is: H.P.

The given series is

$$\frac1{1(1+2)}+\frac1{2(2+4)}+\frac1{4(4+8)}+\cdots$$

$$=\frac13+\frac1{12}+\frac1{48}+\frac1{192}+\cdots$$

This is a G.P. with

$$a=\frac13,\qquad r=\frac14$$

Since

$$|r|<1,$$

the sum to infinity is

$$S_\infty=\frac{a}{1-r}$$

$$=\frac{\frac13}{1-\frac14}$$

$$=\frac{\frac13}{\frac34}$$

$$=\frac13\times\frac43$$

$$=\frac49$$

Hence, the required sum is

$$\boxed{\frac49}$$