The two middle terms in the expansion of {lef | vedclass

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Question

(C) For the expansion of $(x - \frac{1}{x})^{11}$,the index $n = 11$ is odd. Therefore,there are two middle terms given by the terms $T_{\frac{n+1}{2}

Vedclass · Accepted Answer

$-462x$ and $\frac{462}{x}$

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(C) For the expansion of $(x - \frac{1}{x})^{11}$,the index $n = 11$ is odd. Therefore,there are two middle terms given by the terms $T_{\frac{n+1}{2}}$ and $T_{\frac{n+3}{2}}$,which are $T_6$ and $T_7$. The general term is $T_{r+1} = {^{11}C_r} (x)^{11-r} (-x^{-1})^r$. For $T_6$,$r = 5$: $T_6 = {^{11}C_5} (x)^6 (-x^{-1})^5 = 462 \cdot x^6 \cdot (-x^{-5}) = -462x$. For $T_7$,$r = 6$: $T_7 = {^{11}C_6} (x)^5 (-x^{-1})^6 = 462 \cdot x^5 \cdot (x^{-6}) = \frac{462}{x}$.

The two middle terms in the expansion of ${\left( {x - \frac{1}{x}} \right)^{11}}$ are:

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