Q.Using binomial theorem, find the value of $\sqrt[3]{995}$ upto four places of decimals.

The ordinary Binomial Theorem written with $^nC_r=\dfrac{n!}{r!(n-r)!}$ only makes sense when $n$ is a non-negative integer, because $n!$ is undefined otherwise. To expand something like $(1+x)^{-4}$ or $(1-x)^{1/3}$, the theorem must be restated in a form that never needs $n!$ of $n$ itself: for $|x|<1$, $(1+x)^n = 1 + nx + \dfrac{n(n-1)}{2!}x^2 + \dfrac{n(n-1)(n-2)}{3!}x^3 + \cdots + \dfrac{n(n-1)\cdots(n-r+1)}{r!}x^r + \cdots,$ where $n$ can now be any real number — negative, or a fraction, or (as a special case) a positive integer, which is when the series happens to terminate and reduces back to the ordinary finite Binomial Theorem. Two structural differences from the positive-integer case matter a great deal in practice. First, when $n$ is not a positive integer the series never terminates — it has infinitely many terms — and the condition $|x|<1$ is not optional decoration but the exact requirement that makes that infinite sum converge to a finite value; an exam answer is expected to quote 'first three/four terms', not the whole series. Second, to expand a general $(a+b)^n$ (rather than $(1+x)^n$) with $|b|<|a|$, the standard move is to fa …