Progression (Sequence and Series)HardBloom L3

Question

Let $a_{k} = \frac{k}{(k - l)^{4/3} + k^{4/3} + (k + l)^{4/3}}$ and $S_{n} = \sum_{k = 1}^{n}\mspace{2mu} a_{k}$, then

Options

A.$S_{26} > \frac{17}{4}$
B.$S_{26} < \frac{17}{4}$
C.$S_{999} < 50$
D.$S_{999} > 50$

Solution

Sol. [Key Concept: Sequence relations from the definition.]

$$\frac{k}{(k - 1)^{4/3} + k^{4/3} + (k + 1)^{4/3}} < \frac{k}{(k - 1)^{4/3} + ((k - 1)(k + 1))^{2/3} + (k + 1)^{4/3}}$$

$$\begin{matrix} & \ = \frac{\left( (k + 1)^{2/3} - (k - 1)^{2/3} \right)k}{(k + 1)^{2} - (k - 1)^{2}} \\ & \ = \frac{1}{4}\left( (k + 1)^{2/3} - (k - 1)^{2/3} \right) \end{matrix}$$

$$\begin{matrix} \Rightarrow & & \ \ \ \ \ \ \ \ & S_{n} & \ < \frac{1}{4}\sum_{k = 1}^{n}\mspace{2mu}\mspace{2mu}\left( (k + 1)^{2/3} - (k - 1)^{2/3} \right) = \frac{1}{4}\left( (n + 1)^{2/3} + n^{2/3} - 1 \right) \\ \Rightarrow & & \ \ \ \ \ \ \ \ & S_{999} & \ < \frac{1}{4}\left( (1000)^{2/3} + (999)^{2/3} - 1 \right) < \frac{1}{4}(100 + 100 - 1) < 50 \\ & & \ \ \ \ \ \ \ \ & {\text{ }S}_{26} & \ < \frac{1}{4}\left( (27)^{2/3} + (26)^{2/3} - 1 \right) < \frac{1}{4}(9 + 9 - 1) = \frac{17}{4} \end{matrix}$$

✘ (a) $S_{26} < 174$

✔ (b) $S_{26} < \frac{14}{big}{((27)}^{2/3}{+ (26)}^{2/3} - 1big) < 174$

✔ (c) $S_{999} < 50$

✘ (d) $S_{999} < 50$

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