That Is Close 7

Algebra Level 4

∑ p = 1 32 ( 3 p + 2 ) ( ∑ q = 1 10 ( sin ⁡ 2 q π 11 − i cos ⁡ 2 q π 11 ) ) 4 = ? \large \sum_{p =1}^{32} (3p +2) \left( \sum_{q =1}^{10} \left (\sin \frac{2q \pi}{11} - i \cos \frac{2q \pi}{11} \right ) \right)^{4} =\, ?

Clarification: i = − 1 i=\sqrt{-1} .


The answer is 1648.

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Ujjwal Mani Tripathi by Ujjwal Mani Tripathi , 21, India

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1 solution

Sabhrant Sachan
Dec 28, 2016

e − i ( 2 q π 11 ) = cos ⁡ 2 q π 11 − i sin ⁡ 2 q π 11 e^{-i\left(\frac{2q\pi}{11}\right)} = \cos{\dfrac{2q\pi}{11}}-i\sin{\dfrac{2q\pi}{11}}

For my convenience let k = − 2 π 11 k = -\dfrac{2\pi}{11}

S = ∑ q = 1 10 e i q k = e i k + e 2 i k + e 3 i k + ⋯ + e 10 i k S = e i k − e 11 i k 1 − e i k \begin{aligned} S & = \displaystyle \sum_{q=1}^{10} e^{iqk} \\ & = e^{ik}+e^{2ik}+e^{3ik}+\cdots+e^{10ik} \\ S & = \dfrac{e^{ik}-e^{11ik}}{1-e^{ik}} \end{aligned}

e 11 i k = e − 2 π i = cos ⁡ ( − 2 π ) = 1 S = e i k − 1 1 − e i k = − 1 S 4 = 1 e^{11ik} = e^{-2\pi i } = \cos{(-2\pi)} = 1 \\ S = \dfrac{e^{ik}-1}{1-e^{ik}} = -1 \\ S^4 = 1

∑ p = 1 32 3 p + 2 = 3 ⋅ 16 ⋅ 33 + 2 ⋅ 32 = 1648 \displaystyle \sum_{p=1}^{32} 3p+2 = 3\cdot 16\cdot 33+2\cdot 32 = \boxed{1648}

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