If αn=∫0I2cosn x cos nx dx and βn=∫0Isin(nx)sinx dx then for the sequence {αn} and {βn} : {where I = Kπ, K∈N }

αn−1=∫0I2cosn−1x cos(n−1)x dx αn−1=∫0I2cosn−1x(cos nx.cosx+sinnx.sinx)dx αn−1=∫0I/2cosnx. cosnx dx+∫0I/2cosn−1x sinxIBP sin(nx)dx αn−1=αn+αn⇒αn−1=2αn βn={0,if n∈ evenI,if n∈ odd

If αn=∫0I2cosn x cos nx dx and βn=∫0Isin(nx)sinx dx then for the sequence {αn} and {βn} : {where I = Kπ, K∈N }

αn−1=∫0I2cosn−1x cos(n−1)x dx αn−1=∫0I2cosn−1x(cos nx.cosx+sinnx.sinx)dx αn−1=∫0I/2cosnx. cosnx dx+∫0I/2cosn−1x sinxIBP sin(nx)dx αn−1=αn+αn⇒αn−1=2αn βn={0,if n∈ evenI,if n∈ odd

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If $α_{n} = \int_{0}^{\frac{I}{2}} \cos^{n} x \cos n x d x a n d β_{n} = \int_{0}^{I} \frac{\sin (n x)}{\sin x} d x$ then for the sequence ${α_{n}} a n d {β_{n}}$ : {where I = $K π, K \in N$ }

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$α_{1}, α_{2} ....., α_{n}$ are in G.P

$β_{1}, β_{2} ....., β_{n}$ are in G.P

If $I = π t h e n \sum_{i = 1}^{2000} β_{i} = 1000 π$

If $I = π t h e n α_{100} = \frac{π}{2^{100}}$

answer is A, C.

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Detailed Solution

$α_{n - 1} = \int_{0}^{\frac{I}{2}} \cos^{n - 1} x \cos (n - 1) x d x α_{n - 1} = \int_{0}^{\frac{I}{2}} \cos^{n - 1} x (\cos n x . \cos x + \sin n x . \sin x) d x α_{n - 1} = \int_{0}^{I / 2} \cos^{n} x . \cos n x d x + \int_{0}^{I / 2} \frac{\cos^{n - 1} x \sin x}{I B P} \sin (n x) d x α_{n - 1} = α_{n} + α_{n} \Rightarrow α_{n - 1} = 2 α_{n} β_{n} = {\begin{matrix} 0, & i f n \in e v e n \\ I, & i f n \in o d d \end{matrix}$