Geometric Progression

GP with increasing positive terms. $$a_2 \cdot a_3 \cdot a_4 = 64$$.

$$a_2 a_3 a_4 = (a_3/r)(a_3)(a_3 r) = a_3^3 = 64 \Rightarrow a_3 = 4$$.

$$a_1 + a_3 + a_5 = \frac{4}{r^2} + 4 + 4r^2 = \frac{813}{7}$$

Let $$u = r^2$$: $$\frac{4}{u} + 4 + 4u = \frac{813}{7}$$

$$4u^2 + 4u + 4 = \frac{813u}{7}$$

$$28u^2 + 28u + 28 = 813u$$

$$28u^2 - 785u + 28 = 0$$

$$u = \frac{785 \pm \sqrt{785^2 - 4(28)(28)}}{56} = \frac{785 \pm \sqrt{616225 - 3136}}{56} = \frac{785 \pm \sqrt{613089}}{56} = \frac{785 \pm 783}{56}$$

$$u = \frac{1568}{56} = 28$$ or $$u = \frac{2}{56} = \frac{1}{28}$$.

Since GP is increasing: $$r > 1$$, so $$u = r^2 = 28$$.

$$a_3 + a_5 + a_7 = 4 + 4(28) + 4(28^2) = 4 + 112 + 3136 = 3252$$.

The answer is Option 4: 3252.

We are given $$a_1 = 1$$ and the recurrence $$a_{n+1} = \frac{1}{2}a_n + \frac{n^2 - 2n - 1}{n^2(n+1)^2}$$ for $$n \geq 1$$. We need to find $$\left|\sum_{n=1}^{\infty}\left(a_n - \frac{2}{n^2}\right)\right|$$.

Define a new sequence $$b_n = a_n - \frac{2}{n^2}$$ so that the desired sum is $$\left|\sum_{n=1}^{\infty}b_n\right|$$.

Since $$a_{n+1} = b_{n+1} + \frac{2}{(n+1)^2}$$ and $$a_n = b_n + \frac{2}{n^2}$$, substituting into the given recurrence yields $$b_{n+1} + \frac{2}{(n+1)^2} = \frac{1}{2}\Bigl(b_n + \frac{2}{n^2}\Bigr) + \frac{n^2 - 2n - 1}{n^2(n+1)^2},$$ which simplifies to $$b_{n+1} = \frac{b_n}{2} + \frac{1}{n^2} + \frac{n^2 - 2n - 1}{n^2(n+1)^2} - \frac{2}{(n+1)^2}.$$

Combining the terms with common denominator $$n^2(n+1)^2$$ gives $$\frac{1}{n^2} + \frac{n^2 - 2n - 1}{n^2(n+1)^2} - \frac{2}{(n+1)^2} = \frac{(n+1)^2 + (n^2 - 2n - 1) - 2n^2}{n^2(n+1)^2}.$$ Expanding the numerator yields $$(n^2 + 2n + 1) + (n^2 - 2n - 1) - 2n^2 = n^2 + 2n + 1 + n^2 - 2n - 1 - 2n^2 = 0,$$ so that $$b_{n+1} = \frac{b_n}{2}$$.

Thus the sequence $$\{b_n\}$$ is a geometric progression with common ratio $$r = \frac{1}{2}$$ and first term $$b_1 = a_1 - \frac{2}{1^2} = 1 - 2 = -1.$$

The sum of this infinite geometric progression is $$\sum_{n=1}^{\infty} b_n = \frac{b_1}{1 - r} = \frac{-1}{1 - \frac{1}{2}} = \frac{-1}{\frac{1}{2}} = -2,$$ so $$\left|\sum_{n=1}^{\infty} b_n\right| = |-2| = 2.$$

Therefore, the required value is $$\boxed{2}$$.

Let the first three terms of the GP be $$\dfrac{a}{r},a,ar$$

Product of the first three terms = 27

$$\dfrac{a}{r}\times a\times ar=27$$

$$a=3$$

Sum of the first three terms $$=\dfrac{3}{r}+3+3r=3+3\left(r+\dfrac{1}{r}\right)$$

Now if $$r>0$$, then $$r+\dfrac{1}{r}\ge2$$ => $$3\left(r+\dfrac{1}{r}\right)\ge6$$ => $$3+3\left(r+\dfrac{1}{r}\right)\ge9$$

Now if $$r<0$$, then $$r+\dfrac{1}{r}\le-2$$ => $$3\left(r+\dfrac{1}{r}\right)\le-6$$ => $$3+3\left(r+\dfrac{1}{r}\right)\le-3$$

Therefore, the possible values of the sum of the first three terms will be every real number not in the range of $$\left(-3,9\right)$$

Thus, the values will be $$R-\left(-3,9\right)$$

$$a=-3$$ and $$b=9$$

$$a^2+b^2=(-3)^2+9^2=90$$