Question

please solve the question .......​

Answer 1

Answer:

6th sum=

$\sqrt[4]{ 81} + 15 \sqrt[5]{32 } - 8 \sqrt[3]{216} + \sqrt{225}$

$(81) \frac{1}{4 } + 15(32) \frac{1}{5 } - 8(216)\frac{1}{3} + (225) \frac{1}{2}$

$( {3}^{4} ) \frac{1}{4 } + 15( {2}^{5} ) \frac{1}{5} - 8( {6}^{3} ) \frac{1}{3} + (15 ^{2} ) \frac{1}{2}$

3+15(2)-8(6)+15

3+30-48+15

48-48

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