The integral part of (2+1)^6 is

MCQ

The integral part of $(\sqrt{2}+1)^6$ is

A
$98$
B
$96$
✓
$99$
D
$100$

✓

Answer

Correct option: C.

$99$

We have $(1+ x )^{ n }=1+{ }^{ n } C_1(x)+{ }^{ n } C_2(x)^2+\ldots .+( x )^{ n }$
Hence $(\sqrt{2}+1)^6=1+{ }^6 C_1(\sqrt{2})+{ }^6 C_2(\sqrt{2})^2+{ }^6 C_3(\sqrt{2})^3+{ }^6 C_4(\sqrt{2})^4+{ }^6 C_5(\sqrt{2})^5+(\sqrt{2})^6$
$\Rightarrow(\sqrt{2}+1)^6=1+6(\sqrt{2})+15 \times 2+20 \times 2(\sqrt{2})+15 \times 4+6 \times 4(\sqrt{2})+8$
$=99+70 \sqrt{2}$
Hence integral part of $(\sqrt{2}+1)^6=99$`

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