The middle term of (√x-(1/√x))6 is

Question

The middle term of (√x-(1/√x))6 is (A) -20 (B) -1 (C) 1 (D) None of these

Tardigrade Admin · Accepted Answer

Middle term is ((n/2)+1) th term for n even ∴ Middle term =((6/2)+1)=4 th term T4= 6 C3(√x)3(-(1/√x))6-3 =(6 !/3 ! 3 !)(√x)3 ((-1)3/(√x)3) =(-6 × 5 × 4/6)=-20

Q. The middle term of $(x - \frac{1}{x})^{6}$ is

-20

-1

1

None of these

Middle term is $(\frac{n}{2} + 1)$ th term for $n$ even
$∴$ Middle term $= (\frac{6}{2} + 1) = 4$ th term
$T_{4} =^{6} C_{3} (x)^{3} (- \frac{1}{x})^{6 - 3}$
$= \frac{6 !}{3 ! 3 !} (x)^{3} \frac{( - 1 ) ^{3}}{( x ) ^{3}}$
$= \frac{- 6 \times 5 \times 4}{6} = - 20$

Q. The middle term of (x​−x​1​)6 is

-20

-1

1

None of these

Middle term is (2n​+1) th term for n even ∴ Middle term =(26​+1)=4 th term T4​=6C3​(x​)3(−x​1​)6−3 =3!3!6!​(x​)3(x​)3(−1)3​ =6−6×5×4​=−20

Q. The middle term of $(x - \frac{1}{x})^{6}$ is

Middle term is $(\frac{n}{2} + 1)$ th term for $n$ even
$∴$ Middle term $= (\frac{6}{2} + 1) = 4$ th term
$T_{4} =^{6} C_{3} (x)^{3} (- \frac{1}{x})^{6 - 3}$
$= \frac{6 !}{3 ! 3 !} (x)^{3} \frac{( - 1 ) ^{3}}{( x ) ^{3}}$
$= \frac{- 6 \times 5 \times 4}{6} = - 20$