\[{{A}_{n}}=\int\limits_{0}^{\pi /2}{\frac{\sin (2n-1)x}{\sin x}}dx;{{B}_{n}}=\int\limits_{0}^{\pi /2}{{{\left( \frac{\sin nx}{\sin x} \right)}^{2}}dx;}\] For \[n\in N,\] then

\[{{A}_{n+1}}={{A}_{n}},{{B}_{n+1}}-{{B}_{n}}={{A}_{n+1}}\]

\[{{A}_{n+1}}-{{A}_{n}}={{B}_{n+1}}\]

\[{{A}_{n}}=\int\limits_{0}^{\pi /2}{\frac{\sin (2n-1)x}{\sin x}}dx;{{

1.	\[{{A}_{n}}=\int\limits_{0}^{\pi /2}{\frac{\sin (2n-1)x}{\sin x}}dx;{{B}_{n}}=\int\limits_{0}^{\pi /2}{{{\left( \frac{\sin nx}{\sin x} \right)}^{2}}dx;}\] For \[n\in N,\] then
A.	\[{{A}_{n+1}}={{A}_{n}},{{B}_{n+1}}-{{B}_{n}}={{A}_{n+1}}\]
B.	\[{{B}_{n+1}}={{B}_{n}}\]
C.	\[{{A}_{n+1}}-{{A}_{n}}={{B}_{n+1}}\]
D.	None of these
Answer» B. \[{{B}_{n+1}}={{B}_{n}}\]

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\[{{A}_{n}}=\int\limits_{0}^{\pi /2}{\frac{\sin (2n-1)x}{\sin x}}dx;{{B}_{n}}=\int\limits_{0}^{\pi /2}{{{\left( \frac{\sin nx}{\sin x} \right)}^{2}}dx;}\] For \[n\in N,\] then

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