First slide

Binomial theorem for positive integral Index

Question

If [x] denotes the greatest integer less than or equal to x, then $[(6 \sqrt{6} + 14)^{2 n + 1}]$

Moderate

A

is an even integer
B

is an odd integer
C

depends on n
D

none of these

Solution

Let $l + f = (6 \sqrt{6} + 14)^{2 n + 1}$
Assuming, $f' = (6 \sqrt{6} - 14)^{2 n + 1}$ …… (1)
Now, $I + f - f' = (6 \sqrt{6} + 14)^{2 n + 1} - (6 \sqrt{6} - 14)^{2 n + 1}$
$\Rightarrow I + f - f = 2 [^{2 n + 1} C_{1} (6 \sqrt{6})^{2 n} 14^{1} +^{2 n + 1} C_{3} (6 \sqrt{6})^{2 n + 2} (14)^{3} + \dots]$
$\Rightarrow I + f - f' = 2 (Integer) = even$ …… (2)
Now, $0 \leq f < 1$
Also, $0 \leq f - f' < 1$
$∴ 0 \leq f - f' < 0 \Rightarrow f - f' = 0$
Substituting respective values in (2), we get
I = even integer

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