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r-1-n-3-r-1-sin-3-3-r-

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Question Number 185423 by alcohol last updated on 21/Jan/23
Σ_(r=1) ^n 3^(r−1) sin^3 ((θ/3^r )) = ?
∑nr=13r−1sin3(θ3r)=?
Answered by witcher3 last updated on 21/Jan/23
sin^3 (x)=−(1/4)(((e^(3ix) −e^(−3ix) −3e^(ix) +3e^(−ix) )/(2i))) =−(1/4)sin(3x)+(3/4)sin(x) S=Σ_(r=1) ^n 3^(r−1) sin^3 ((θ/3^r ))=−(1/4)(Σ_(r=1) ^n 3^(r−1) sin((θ/3^(r−1) ))−Σ_(r=1) ^n 3^r sin((θ/3^r ))) =−(1/4)(sin(θ)−3^n sin((θ/3^n )))
sin3(x)=−14(e3ix−e−3ix−3eix+3e−ix2i)=−14sin(3x)+34sin(x)S=∑nr=13r−1sin3(θ3r)=−14(∑nr=13r−1sin(θ3r−1)−∑nr=13rsin(θ3r))=−14(sin(θ)−3nsin(θ3n))

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