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Q.

If the $21^{st}$ and $22^{nd}$ terms in the expansion ${(1 + x)}^{44}$ are equal then $x =$

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answer is 0.87.

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Detailed Solution

$\begin{array}{rcl} In {(1 + x)}^{44} T_{21} & = & T_{22} \\ {}^{44}{C_{20}} x^{20} & = & {}^{44}{C_{21}} x^{21} \\ \frac{1}{x} & = & \frac{{}^{44}{C_{21}}}{{}^{44}{C_{20}}} = \frac{44 - 21 + 1}{21} \\ \frac{1}{x} & = & \frac{24}{21} \\ x & = & 0.87 \end{array}$

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