Suppose [x] denote the greatest integer ≤x and n∈N, then limn→∞ nCox2+ nC1x2+⋯+ nCnx22n−2 is equal to

A
$\frac{1}{2} x^{2}$
B
$x^{2}$
C
$2 x^{2}$
D
$4 x^{2}$

Solution:

We know $x - 1 < [x] \leq x \forall x \in R$ , there fore $^{n} C_{k} x^{2} - 1 < [^{n} C_{k} x^{2}] \leq^{n} C_{k} x^{2}$

$\begin{array}{l} \Rightarrow \sum_{k = 0}^{n} (^{n} C_{k} x^{2} - 1) < \sum_{k = 0}^{n} [^{n} C_{k} x^{2}] \leq \sum_{k = 0}^{n}^{n} C_{k} x^{2} \\ \Rightarrow \frac{x^{2} (2^{n}) - (n + 1)}{2^{n} - 2} < \frac{1}{2^{n} - 2} \sum_{k = 0}^{n} [^{n} C_{k} x^{2}] \\ \leq \frac{x^{2} (2^{n})}{2^{n} - 2} \end{array}$

Taking limit as $n \to \infty$ ,and using sandwich theorem, we get

$lim_{n \to \infty} \frac{1}{2^{n} - 2} \sum_{k = 0}^{n} [^{n} C_{k} x^{2}] = x^{2}$

Suppose [x] denote the greatest integer $\leq x$ and $n \in N$ , then
$lim_{n \to \infty} \frac{[^{n} C_{o} x^{2}] + [^{n} C_{1} x^{2}] + \dots + [^{n} C_{n} x^{2}]}{2^{n} - 2}$ is equal to

Solution:

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