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`IfA_n=int_0^(pi/2)(sin(2n-1)x)/(sinx)dx ,b_n=int_0^(pi/2)((sinn x)/(sinx))^2dxforn in N ,`Then`A_(n+1)=A_n`(b) `B_(n+1)=B_n``A_(n+1)-A_n=B_(n+1)`(d) `B_(n+1)-B_n=A_(n+1)`A. `A_(n+1)=A_(n)`B. `B_(n+1)=B_(n)`C. `A_(n+1)-A_(n)=B_(n+1)`D. `B_(n+1)-B_(n)=A_(n+1)` |
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Answer» Correct Answer - A::D `A_(n+1)-A_(n)=int_(0)^(pi//2)(sin(2n+1)xsin(2n-1)x)/(sinx) dx` `=int_(0)^(pi//2)2 cos 2nx dx=0` or `A_(n+1)=A_(n)` `B_(n+1)-B_(n)=int_(0)^(pi//2)(sin^(2)(n+1)x-sin^(2)nx)/(sin^(2)x)dx` `=int_(0)^(pi//2) (sin(2n+1)x)/(sinx)dx=A_(n+1)` |
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