(ii). The fith term in the expansion of \( \left(1-\frac{1}{x}\right)^{10} \) is : (a) \( \frac{210}{x^{4}} \) (b) \( \frac{x^{4}}{210} \) (c) \( \frac{1}{x^{4}} \) (d) None \( \square \times \square \) (iv). The middle term/s in the expansion of \( \left(x+\frac{1}{x}\right)^{10} \) is : (a) 504 (b) 252 (c) \( \frac{1}{x^{4}} \) (d) None (v). The value of \( i^{6} \), (where \( i=\sqrt{-1} \) ): (a) 1 (b) \( i \) (c) -1 (d) None (vi). The value of \( i^{6}+\omega^{6} \) : (a) 1 (b) 2 (c), 0 (d) None
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The fifth term in the expansion of \( \left(1-\frac{1}{x}\right)^{10} \) can be found using the binomial theorem, which states that the \( r^{th} \) term is given by \( T_r = \binom{n}{r-1} a^{n-(r-1)} b^{r-1} \). Here, with \( n=10 \), \( a=1 \), and \( b=-\frac{1}{x} \), the fifth term \( T_5 \) is \( \binom{10}{4} (1)^{6} \left(-\frac{1}{x}\right)^{4} = -\frac{210}{x^{4}} \), so the correct answer is (a) \( \frac{210}{x^{4}} \). For the middle term in the expansion of \( \left(x+\frac{1}{x}\right)^{10} \), we need to recognize that this binomial expands to \( 2n \) terms when \( n \) is even. In this case, since \( n=10 \), the middle terms are \( T_5 \) and \( T_6 \), both of which can be computed. Specifically, using \( T_k = \binom{n}{k} a^{n-k} b^{k} \), we get \( T_5 = \binom{10}{5} x^5 \left(\frac{1}{x}\right)^5 = \binom{10}{5} = 252 \) and hence, (b) 252 is the answer. The final values can also be calculated based on \( i^{6} \). Utilizing the property \( i^2 = -1 \), we find \( i^{6} = (i^2)^{3} = (-1)^{3} = -1 \), so the answer is (c) -1. For \( \omega^{6} \), where \( \omega \) is a cube root of unity, it satisfies \( \omega^3 = 1 \), hence \( \omega^{6} = (\omega^3)^2 = 1 \). Therefore, \( i^6 + \omega^6 = -1 + 1 = 0 \), leaving us with (c) 0.
