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  4. If the 9 text th term in the expansion of ((1/x2)+(x/2) log 2 x)12 is equal to 495 , then sum of all possible values of x is

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Q. If the $9^{\text {th }}$ term in the expansion of $\left(\frac{1}{x^2}+\frac{x}{2} \log _2 x\right)^{12}$ is equal to 495 , then sum of all possible values of $x$ is

Binomial Theorem

Solution:

$ T_9={ }^{12} C_8\left(\frac{1}{x^2}\right)^4\left(\frac{x \log _2 x}{2}\right)^8={ }^{12} C_4 \frac{\left(\log _2 x\right)^8}{2^8}=495$.
Since, ${ }^{12} C_4=495,\left(\log _2 x\right)^8=2^8$ and $\log _2 x= \pm 2$ giving $x=4, \frac{1}{4}$