Q175 / 270 · NDA 2019 (Phase 2) · Mathematics

If the middle term in the expansion of $\left(x^2+\frac{1}{x}\right)^{2 x}$ $184756 x^{10}$, then what is the value of $n$ ?

  • Correct answer
    A
    10
  • B
    8
  • C
    5
  • D
    4

Solution

$\left(x^2+\frac{1}{x}\right)^{2 n}$ Its index is even $\begin{aligned} & \therefore \quad \text { Middle term }=T_{n+1} \\ & T_{n+1}={ }^{2 n} C_n\left(x^2\right)^n\left(\frac{1}{x}\right)^n \\ & ={ }^{2 n} C_n x^n=184756 x^{10} \end{aligned}$ $\therefore$ comparing power of $x$ we get $n=10$

Hence, the correct answer is option: 'A'.

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