mathbb{P} P=5 D^{1} cdot frac{(1+i)^{2} i}{(1+1)^{N}-1}

Question

Accepted Answer

Step:1 ：Given values: (D = 1), (i = 1), and (N = 5)Step:2 ：Plug these values into the formula: (P = 5 cdot 1^{1} cdot frac{(1+1)^{2} cdot 1}{(1+1)^{5}-1})Step:3 ：Simplify the expression: (P = 5 cdot frac{2^{2}}{2^{5}-1})Step:4 ：Calculate the value of P: (P = 5 cdot frac{4}{31})Step:5 ：(boxed{P approx 0.6451612903225806})

Answer

Step: 1 ：Given values: (D = 1), (i = 1), and (N = 5)Step: 2 ：Plug these values into the formula: (P = 5 cdot 1^{1} cdot frac{(1+1)^{2} cdot 1}{(1+1)^{5}-1})Step: 3 ：Simplify the expression: (P = 5 cdot frac{2^{2}}{2^{5}-1})Step: 4 ：Calculate the value of P: (P = 5 cdot frac{4}{31})Step: 5 ：(boxed{P approx 0.6451612903225806})

$\mathbb{P} P=5 D^{1} \cdot \frac{(1+i)^{2} i}{(1+1)^{N}-1}$

Answer

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