If I is integral part of (2+3)n and/is its fractional part. Then (I+f)(1−f) is

Given (2+3)n=I+f where l is integer and 0≤f<1. We note that (2+3)(2−3)=1 so let us assume thatvF=(2−3)nClearly 0<F<1. Now, I+f+F=(2+3)n+(2−3)n=2 nC02n+nC22n−2⋅3+nC4⋅2n−4⋅32+………..=2×Integer= even Integer∵I+f+F is integer ⇒f+Fmust be integer.∴0≤f<1 and 0<F<1⇒0<f+F<2⇒f+F=1⇒F=1−f∴(I+f)(1−f)=(I+f)F=(2+3)n(2−3)n=1

If I is integral part of (2+3)n and/is its fractional part. Then (I+f)(1−f) is

Given (2+3)n=I+f where l is integer and 0≤f<1. We note that (2+3)(2−3)=1 so let us assume thatvF=(2−3)nClearly 0<F<1. Now, I+f+F=(2+3)n+(2−3)n=2 nC02n+nC22n−2⋅3+nC4⋅2n−4⋅32+………..=2×Integer= even Integer∵I+f+F is integer ⇒f+Fmust be integer.∴0≤f<1 and 0<F<1⇒0<f+F<2⇒f+F=1⇒F=1−f∴(I+f)(1−f)=(I+f)F=(2+3)n(2−3)n=1

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If I is integral part of $(2 + \sqrt{3})^{n}$ and/is its fractional part. Then $(I + f) (1 - f)$ is

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$I + 1$

$2^{n}$

answer is B.

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Detailed Solution

Given $(2 + \sqrt{3})^{n} = I + f$ where l is integer and $0 \leq f < 1 . We note that (2 + \sqrt{3}) (2 - \sqrt{3}) = 1$ so let us assume that

$v F = (2 - \sqrt{3})^{n}$ Clearly $0 < F < 1 . Now,$

$\begin{array}{l} I + f + F = (2 + \sqrt{3})^{n} + (2 - \sqrt{3})^{n} \\ = 2 [^{n} C_{0} 2^{n} +^{n} C_{2} 2^{n - 2} \cdot 3 +^{n} C_{4} \cdot 2^{n - 4} \cdot 3^{2} + \dots \dots \dots . .] \end{array}$

$= 2 \times$ Integer= even Integer

$∵ I + f + F$ is integer $\Rightarrow f + F$ must be integer.

$\begin{array}{l} ∴ 0 \leq f < 1 and 0 < F < 1 \Rightarrow 0 < f + F < 2 \\ \Rightarrow f + F = 1 \Rightarrow F = 1 - f \\ ∴ (I + f) (1 - f) = (I + f) F = (2 + \sqrt{3})^{n} (2 - \sqrt{3})^{n} = 1 \end{array}$