∑k=1513+23+……+k31+3+5+……+(2k-1)=

∑k=15 13+23+……+k31+3+5+……+(2k-1) =∑k=15 k2(k+1)24k2=∑k=15 (k+1)24 =14 22+32+42+52+62 =14 4+9+16+25+36 =904=22.5

∑k=1513+23+……+k31+3+5+……+(2k-1)=

∑k=15 13+23+……+k31+3+5+……+(2k-1) =∑k=15 k2(k+1)24k2=∑k=15 (k+1)24 =14 22+32+42+52+62 =14 4+9+16+25+36 =904=22.5

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$\sum_{k = 1}^{5} \frac{1^{3} + 2^{3} + \dots \dots + k^{3}}{1 + 3 + 5 + \dots \dots + (2 k - 1)} =$

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22.5

24.5

28.5

32. 5

answer is A.

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Detailed Solution

$\begin{array}{rcl} \sum_{k = 1}^{5} \frac{1^{3} + 2^{3} + \dots \dots + k^{3}}{1 + 3 + 5 + \dots \dots + (2 k - 1)} \\ = & \sum_{k = 1}^{5} \frac{k^{2} {(k + 1)}^{2}}{4 k^{2}} = \sum_{k = 1}^{5} \frac{{(k + 1)}^{2}}{4} \\ = & \frac{1}{4} (2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}) \\ = & \frac{1}{4} (4 + 9 + 16 + 25 + 36) \\ = & \frac{90}{4} = 22.5 \end{array}$