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CBSE Class 11-commerce Answered

Question no. 21
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Asked by Payal | 22 Feb, 2018, 12:14: PM
answered-by-expert Expert Answer
begin mathsize 16px style Using space straight T subscript straight r plus 1 end subscript equals straight C presuperscript straight n subscript straight r space straight x to the power of straight n minus straight r end exponent space straight a to the power of straight r space straight n equals 11 space you space will space get straight x to the power of 22 minus 3 straight r end exponent equals straight x to the power of 7 22 minus 3 straight r equals 7 straight r equals 5 then space find space coefficient space of space straight x to the power of 7 space in space open parentheses ax squared plus 1 over bx close parentheses to the power of 11 which space is space straight C presuperscript 11 subscript 5 space open parentheses straight a to the power of 6 over straight b to the power of 5 close parentheses Let space straight x to the power of negative 7 space end exponent space occurs space in space open parentheses straight r plus 1 close parentheses to the power of th space term hence space solving space using space straight T subscript straight r plus 1 end subscript equals straight C presuperscript straight n subscript straight r space straight x to the power of straight n minus straight r end exponent space straight a to the power of straight r space you space get space straight r equals 6 Also space find space coefficient space of space straight x to the power of negative 7 space end exponent Equating space coefficient space of space straight x to the power of 7 space in space space open parentheses ax squared plus 1 over bx close parentheses to the power of 11 equals coefficient space of space straight x to the power of negative 7 end exponent space in space space open parentheses ax minus 1 over bx squared close parentheses to the power of 11 you space will space get space ab equals 1 end style
Answered by Sneha shidid | 22 Feb, 2018, 01:27: PM

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