MCQ
$1 + n\left( {1 - \frac{1}{x}} \right) + \frac{{n(n + 1)}}{{2!}}{\rm{ }}{\left( {1 - \frac{1}{x}} \right)^2} + .....\infty $ = . . . .
  • ${x^n}$
  • B
    ${x^{ - n}}$
  • C
    ${\left( {1 - \frac{1}{x}} \right)^n}$
  • D
    એકપણ નહીં.

Answer

Correct option: A.
${x^n}$
(a) We have

${(1 + x)^n} = {\,^n}{C_0} + {\,^n}{C_1}x + {\,^n}{C_2}{x^2} + .....\infty $

If $x$ is replace by $ - \left( {1 - \frac{1}{x}} \right)$ and $n$ is $ - n$, then expression becomes ${\left[ {1 - \left( {1 - \frac{1}{x}} \right)} \right]^{ - n}}.$

$ = 1 + ( - n)\,\left[ { - \left( {1 - \frac{1}{x}} \right)} \right] + \frac{{( - n)( - n - 1)}}{{2!}}{\left[ { - \left( {1 - \frac{1}{x}} \right)} \right]^2} + ...$

or ${x^n} = 1 + n\left( {1 - \frac{1}{x}} \right) + \frac{{n(n + 1)}}{{2!}}{\left( {1 - \frac{1}{x}} \right)^2} + ....$

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