The coefficient of the (r +1)th term of `(x +(1)/(x))^(20)` , when
expanded in the descending power of x , is equal to the
coefficient of the 6th term of `(x^(2) + 2 + (1)/(x^(2))) ` when
expanded in ascending power of x . The value of r is

Options

A. 5
B. 6
C. 14
D. 15

Answer

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Explanation

Now `(x + (1)/(x))^(20) = ""^(20)C_(0) x^(20) + ""^(20)C_(1)x^(18) + ""^(20)C_(2) x^(16) +""^(20)C_(3) x^(14) + …`
` + ""^(20)C_(9) x^(2) + ""^(20)C_(10) + ""^(20)C_(11) x^(-2) + …+ ""^(20)C_(20) x^(-20)`
` T_(r+1) = ""^(20)C_(r) . X^(20 - 2r) ` ...(i)
and `(x^(2) + 2 + (1)/(x^(2)))^(10) = =((1)/(x) + x)^(20) `
` = ""^(20)C_(0) x^(-20) + ""^(20)C_(1) x^(-18) + ""^(20)C_(2) x^(-16) `
` + ...+ ""^(20)C_(10) + ""^(20)C_(11)x^(2) + ""^(20)C_(12) x^(4)`
` + ...+ ""^(20)C_(20) x^(20)`
` therefore T_(6) = t_(5+1) = ""^(20)C_(5) x^(-10)` ...(ii)
According to the question , `""^(20)C_(r) = ""^(20)C_(5) `
` therefore r= 5 or 20 = r + 5 rArr Sr = 5 , 15 `