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  4. displaystyle ∑5k=1 (13+ 23 + .... +k3/1+3+5+.....+(2k-1)) =

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Q. $ \displaystyle \sum^{5}_{k=1} \frac{1^{3}+ 2^{3} + .... +k^{3}}{1+3+5+.....+\left(2k-1\right)} = $

Principle of Mathematical Induction

Solution:

$ \displaystyle \sum^{5}_{k=1} \frac{(k+1)^2}{4} = \frac{4}{4}+\frac{9}{4}+\frac{16}{4}+\frac{25}{4} + \frac{36}{4} = \frac{90}{4} = 22.5 $