Sequence

We are given $$f(1) + 2f(2) + 3f(3) + \ldots + xf(x) = x(x+1)f(x)$$ for $$x \geq 2$$, with $$f(1) = 1$$. Defining $$S(x) = \sum_{k=1}^{x} kf(k)$$ shows that $$S(x) = x(x+1)f(x)$$ for $$x \geq 2$$.

To find $$f(2)$$, observe that $$S(2) = f(1) + 2f(2) = 1 + 2f(2)$$ and $$S(2) = 2 \cdot 3 \cdot f(2) = 6f(2)$$, hence $$1 + 2f(2) = 6f(2) \implies f(2) = \frac{1}{4}$$.

For $$x \geq 3$$, since $$S(x) - S(x-1) = xf(x)$$, with $$S(x) = x(x+1)f(x)$$ and $$S(x-1) = (x-1)xf(x-1)$$, one has $$x(x+1)f(x) - (x-1)xf(x-1) = xf(x)$$. Simplifying gives $$x^2 f(x) = x(x-1)f(x-1)$$ and hence $$xf(x) = (x-1)f(x-1)$$.

Iterating this recurrence yields $$xf(x) = (x-1)f(x-1) = \ldots = 2f(2) = \frac{1}{2}$$, so that $$f(x) = \frac{1}{2x}$$ for all $$x \geq 2$$.

As a verification, $$f(3) = \frac{1}{6}$$, and $$S(3) = 1 + 2 \cdot \frac{1}{4} + 3 \cdot \frac{1}{6} = 1 + \frac{1}{2} + \frac{1}{2} = 2$$, while $$3 \cdot 4 \cdot f(3) = 12 \cdot \frac{1}{6} = 2$$.

Finally, $$\frac{1}{f(2022)} + \frac{1}{f(2028)} = 2 \times 2022 + 2 \times 2028 = 4044 + 4056 = \boxed{8100}$$.

The correct answer is Option D.

Given,

$$a_{n+1}=a_n^2-2a_n+2$$

Rewrite as

$$a_{n+1}-1=a_n^2-2a_n+1$$

$$a_{n+1}-1=(a_n-1)^2$$

Let $$b_n=a_n-1$$

Then $$b_{n+1}=b_n^2$$

Since $$a_1=3$$

we have $$b_1=2$$

Therefore,

$$b_2=2^2=2^{2}$$

$$b_3=(2^2)^2=2^{2^2}$$

$$b_4=(2^{2^2})^2=2^{2^3}$$

Continuing,

$$b_n=2^{2^{\,n-1}}$$

Hence,

$$a_n=1+2^{2^{\,n-1}}$$

Now,

$$1-\frac1{a_n}=\frac{a_n-1}{a_n}$$

$$=\frac{2^{2^{\,n-1}}}{1+2^{2^{\,n-1}}}$$

Let

$$m_n=2^{2^{\,n-1}}$$ Then $$m_{n+1}=m_n^2$$

and

$$1-\frac1{a_n}=\frac{m_n}{1+m_n}$$

Therefore,

$$P_N=\prod_{n=1}^{N}\frac{m_n}{1+m_n}$$

Substituting $$m_n=2^{2^{\,n-1}}$$ gives

$$P_N=\frac2{1+2}\cdot\frac{2^2}{1+2^2}\cdot\frac{2^4}{1+2^4}\cdots\frac{2^{2^{N-1}}}{1+2^{2^{N-1}}}$$

Now use the identity

$$x^{2^N}-1=(x-1)(x+1)(x^2+1)(x^4+1)\cdots(x^{2^{N-1}}+1)$$

Putting $$x=2$$ we get $$2^{2^N}-1=(2-1)(2+1)(2^2+1)(2^4+1)\cdots(2^{2^{N-1}}+1)$$

Since $$2-1=1$$ $$\prod_{k=0}^{N-1}(1+2^{2^k})=2^{2^N}-1$$

Therefore,

$$P_N=\frac{2^{1+2+4+\cdots+2^{N-1}}}{2^{2^N}-1}$$

The exponent sum is

$$1+2+4+\cdots+2^{N-1}=2^N-1$$

Hence,

$$P_N=\frac{2^{2^N-1}}{2^{2^N}-1}$$

Taking limit as

$$N\to\infty$$

$$P=\lim_{N\to\infty}\frac{2^{2^N-1}}{2^{2^N}-1}$$

Divide numerator and denominator by

$$2^{2^N}$$

$$P=\lim_{N\to\infty}\frac{\frac12}{1-\frac1{2^{2^N}}}$$

$$=\frac{\frac12}{1-0}$$

$$=\frac12$$

Let's write out the first few terms to find the pattern:
• $$a_1 = 2$$
• $$a_2 = 2 + 3 = 5$$ (since $$n=2$$ is even)
• $$a_3 = 5 \times 2 = 10$$ (since $$n=3$$ is odd)
• $$a_4 = 10 + 3 = 13$$
• $$a_5 = 13 \times 2 = 26$$
• $$a_6 = 26 + 3 = 29$$
Notice a shortcut by looking at the even-indexed terms ($$a_2, a_4, a_6, \dots$$):
• $$a_2 = 5$$
• $$a_4 = 13$$
• $$a_6 = 29$$
The pattern for the even terms is $$a_{2k} = 2 \cdot a_{2k-2} + 3$$.
Let's continue the even chain up to $$a_{10}$$:
• $$a_6 = 29$$
• $$a_8 = (29 \times 2) + 3 = 61$$
• $$a_{10} = (61 \times 2) + 3 = 125$$ ... Wait, let's re-verify the strict definition rules.
Let's trace directly from $$a_6$$:
• $$a_6 = 29$$
• $$a_7 = 29 \times 2 = 58$$
• $$a_8 = 58 + 3 = 61$$
• $$a_9 = 61 \times 2 = 122$$
• $$a_{10} = 122 + 3 = \mathbf{125}$$