First slide

Binomial theorem for positive integral Index

Question

If $[]$ denotes the greatest integer function, then $[(\sqrt{2} + 1)^{6}]$ is equal

Moderate

A

$199$
B

$198$
C

$197$
D

$196$

Solution

Let $(\sqrt{2} + 1)^{6} = R + f$ where $R$ is an integer and $0 \leq f < 1 .$
Note that $R = [(\sqrt{2} + 1)^{6}]$
Also, $0 < \sqrt{2} - 1 < 1 \Rightarrow 0 < F = (\sqrt{2} - 1)^{6} < 1$
We have, $R + f + F = (\sqrt{2} + 1)^{6} + (\sqrt{2} - 1)^{6}$
$\begin{array}{l} = 2 [^{6} C_{0} (\sqrt{2})^{6} +^{6} C_{2} (\sqrt{2})^{4} +^{6} C_{4} (\sqrt{2})^{2} +^{6} C_{6}] \\ = 2 [2^{3} + 15 (2^{2}) + 15 (2) + 1] = 198 \end{array}$
$\Rightarrow f + F = 198 - R$ is an integer.
As $0 \leq f < 1$ and $0 < F < 1, 0 < f + F < 2 .$
But the only integer between $0$ and $2$ is $1$ . Therefore
$f + F = 1$
Thus, $1 = 198 - R \Rightarrow R = 197$

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