Find-the-simplest-form-of-k-1-n-2-k-sin-2-2kpi-3-1-4-

Question Number 21825 by Joel577 last updated on 05/Oct/17

Find the simplest form of

\underset{k = 1}{\sum^{n}} 2^{k} [\sin^{2} (\frac{2 k π}{3}) + \frac{1}{4}]

Commented by sma3l2996 last updated on 05/Oct/17

w e h a v e : s i n (\frac{2 k π}{3}) = s i n (\frac{2 π}{3})

s o \underset{k = 1}{\sum^{n}} 2^{k} ({s i n}^{2} (\frac{2 k π}{3}) + \frac{1}{4}) = \underset{k = 1}{\sum^{n}} 2^{k} ({s i n}^{2} \frac{2 π}{3} + \frac{1}{4})

= \underset{k = 1}{\sum^{n}} 2^{k} ({(\frac{\sqrt{3}}{2})}^{2} + \frac{1}{4}) = \underset{k = 1}{\sum^{n}} 2^{k} (\frac{3}{4} + \frac{1}{4})

= \underset{k = 1}{\sum^{n}} 2^{k}

l e t a_{k} = 2^{k} = a_{1} q^{k - 1} = 2 \times 2^{k - 1}

s o q = a_{1} = 2

l e t S_{n} = a_{1} + a_{2} + \dots + a_{n} = \frac{a_{1} (q^{n} - 1)}{q - 1}

S_{n} = \underset{k = 0}{\sum^{n}} 2^{k} = \frac{2 (2^{n} - 1)}{2 - 1} = 2 (2^{n} - 1)

s o

\underset{k = 0}{\sum^{n}} 2^{k} ({s i n}^{2} (\frac{2 k π}{3}) + \frac{1}{4}) = 2^{n + 1} - 2

Commented by Joel577 last updated on 05/Oct/17

but if k = 3, then \sin^{2} (\frac{6 π}{3}) \neq \sin^{2} (\frac{2 π}{3})

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