```latex 23. \text{Let } a_k = \frac{k}{(k-1)^{4/3} + k^{4/3} + (k+1)^{4/3}} \text{ and } S_n = \sum_{k=1}^{n} a_k, \text{ then} \text{(A) } S_{26} \geq \frac{17}{4} \quad \text{(B) } S_{26} < \frac{17}{4} \quad \text{(C) } S_{999} < 50 \quad \text{(D) } S_{999} > 50 ```

1. Spot the general behaviour of $a_k$

For large $k$,

$(k-1)^{4/3},\; k^{4/3},\; (k+1)^{4/3}\approx k^{4/3}$

so the denominator is a bit more than $3k^{4/3}$. Therefore

What does the question say?

We have lots of tiny numbers $a_k$. Each one looks like this

$a_k = \frac{k}{(k-1)^{4/3} + k^{4/3} + (k+1)^{4/3}}$

If we keep adding them from $k=1$ upward, we get a running total that we call $S_n$ (the first $n$ of them added together).
The question gives us four statements about two particular running totals:

• $S_{26}$ – the first $26$ terms
• $S_{999}$ – the first $999$ terms

and it asks: which statements are true?

To decide, we have to know roughly how big these sums are.

Why does the sum behave the way it does?

Look at just one term for large $k$:

$a_k \;\approx\; \frac{k}{3\,k^{4/3}} \;=\; \frac{1}{3}\,k^{-1/3}$

Think of $k^{-1/3}$ as a slowly shrinking block. If you pile them up, the total height grows like the area under a curve $y = k^{-1/3}$.

Area idea:

• Area from $1$ to $n$ under $y = k^{-1/3}$ is about $(3/2)\,n^{2/3}$.
• Multiply by $1/3$ (because of the $1/3$ in front) and you find $S_n$ is roughly $(1/2)\,n^{2/3}$.

So:

• When $n = 26$ you expect about $(1/2)\,26^{2/3}\;\approx\;4.2$.
• When $n = 999$ you expect about $(1/2)\,999^{2/3}\;\approx\;50$.

Using a little "area over–area under" comparison (integral bounds) makes those guesses into inequalities that prove

• $S_{26} < 4.25$
• $S_{999} < 50$

Hence the correct options are (B) and (C).

Step-by-Step Solution

1. Upper bound each term

   a_k &= \frac{k}{(k-1)^{4/3}+k^{4/3}+(k+1)^{4/3}} \\ &< \frac{k}{3k^{4/3}} = \frac{1}{3}\,k^{-1/3}. \end{aligned}$$

2. Use the integral comparison

   Let $f(x)=x^{-1/3}$ (positive, decreasing). Then

   $\sum_{k=1}^{n}f(k) \;<\; f(1) + \int_{1}^{n} f(x)\,dx.$

   Compute the integral

   $\int_{1}^{n} x^{-1/3}\,dx = \frac{3}{2}(n^{2/3}-1).$

   Hence

   $S_n < \frac{1}{3} + \frac{1}{3}\cdot\frac{3}{2}(n^{2/3}-1) = \frac{1}{2}n^{2/3} - \frac{1}{6}.$

3. Evaluate the bound for $n=26$

   $26^{2/3}\;\approx\;8.773\quad\Rightarrow\quad S_{26}<\frac{1}{2}\times8.773-\frac{1}{6}\approx4.2198.$

   Since $4.2198<4.25=\dfrac{17}{4}$,

   $S_{26}<\frac{17}{4}.$ So (B) is correct and (A) is not.

4. Evaluate the bound for $n=999$

   $999^{2/3}\;\approx\;99.93\quad\Rightarrow\quad S_{999}<\frac{1}{2}\times99.93-\frac{1}{6}\approx49.8.$

   Because $49.8<50$,

   $S_{999}<50.$ Thus (C) is correct and (D) is not.

5. Final Answer

   The true statements are (B) and (C).