First slide

Binomial theorem for positive integral Index

Question

If $(4 + \sqrt{15})^{n} = I + f$ , where n is an odd natural number, I is an integer and 0 < f < 1, then

Moderate

A

I is an odd integer
B

I is an even integer
C

$(I + f) (1 - f) = 1$
D

$1 - f = (4 - \sqrt{15})^{n}$

Solution

$I + f = (4 + \sqrt{15})^{n}$
Let $f' = (4 + \sqrt{15})^{n}$ . Then 0 < f' < 1
$I + f =^{n} C_{0} 4^{n} +^{n} C_{1} 4^{n - 1} \sqrt{15} +^{n} C_{2} 4^{n - 2} 15 +^{n} C_{3} 4^{n - 3} (\sqrt{15})^{3} + \dots$
$f' =^{n} C_{0} 4^{n} -^{n} C_{1} 4^{n - 1} \sqrt{15} +^{n} C_{2} 4^{n - 2} \cdot 15 -^{n} C_{3} 4^{n - 3} (\sqrt{15})^{3} + \dots$
$∴ I + f + f' = 2 (^{n} C_{0} 4^{n} +^{n} C_{2} 4^{n - 2} \times 15 + \dots) =$ even integer
$∵ 0 < f + f' < 2 \Rightarrow f + f' = 1 \Rightarrow 1 - f = f'$
Thus, I is an odd integer. Now,
$1 - f = f' = (4 - \sqrt{15})^{n}$
$(I + f) (1 - f) = (I + f) f' = 1$

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