Q200 / 270 · NDA 2022 (Phase 1) · Mathematics

In the expansion of $\left(x+\frac{1}{x}\right)^{2 n}$, what is the $(n+1)$ th term from the end (when arranged in descending powers of $x$ )?

  • A
    $C(2 n, n) x$
  • B
    $C(2 n, n-1) x$
  • Correct answer
    C
    $C(2 n, n)$
  • D
    $C(2 n, n-1)$

Solution

$(x+1)^{\text {th }}$ term from the end of expression $\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{2 \mathrm{n}}=(\mathrm{n}+1)^{\text {th }}$ term from begining of expression $\left(\frac{1}{x}+x\right)^{2 n}$ $\therefore(\mathrm{n}+1)^{\text {lh }}$ term of $\left(\frac{1}{\mathrm{x}}+\mathrm{x}\right)^{2 \mathrm{n}}$ is $\mathrm{T}_{\mathrm{n}+1}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}\left(\frac{1}{\mathrm{x}}\right)^{\mathrm{n}} \mathrm{x}^{\mathrm{n}}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}$

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

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