First slide

Binomial theorem for positive integral Index

Question

If $p = (8 + 3 \sqrt{7})^{n}$ and $f = p - [p]$ , where [.] denotes the greatest integer function, then the value of p(1 - f) is equal to

Moderate

A

1
B

2
C

2ⁿ
D

2²ⁿ

Solution

$\begin{array}{l} p = (8 + 3 \sqrt{7})^{n} =^{n} C_{0} 8^{n} +^{n} C_{1} 8^{n - 1} (3 \sqrt{7}) + \dots \\ 49 < 63 < 64 \Rightarrow 7 < \sqrt{63} < 8 \Rightarrow - 8 < - 3 \sqrt{7} < - 7 \Rightarrow 0 < 8 - 3 \sqrt{7} < 1 \\ 0^{n} < {(8 - 3 \sqrt{7})}^{n} < 1^{n} \Rightarrow 0 < {(8 - 3 \sqrt{7})}^{n} < 1 \\ Let p_{1} = (8 - 3 \sqrt{7})^{n} \\ p_{1} = (8 - 3 \sqrt{7})^{n} =^{n} C_{0} 8^{n} -^{n} C_{1} 8^{n - 1} (3 \sqrt{7}) + \dots \\ p + p_{1} = 2 (^{n} C_{0} 8^{n} +^{n} C_{2} 8^{n - 2} (3 \sqrt{7})^{2} + \dots) = even integer \\ p_{1} clearly belongs to (0, 1) \\ \Rightarrow [p] + f + p_{1} = even integer \\ \Rightarrow f + p_{1} = integer \\ f \in (0, 1), p_{1} \in (0, 1) \\ \Rightarrow f + p \in (0, 2) \\ \Rightarrow f + p_{1} = 1 \\ \Rightarrow p_{1} = 1 - f \end{array}$
Now $p (1 - f) = {pp}_{1} = {[(8 + 3 \sqrt{7})^{n} (8 - 3 \sqrt{7})]}^{n} = 1$

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