Binomial Theorem
Grade 11
Question:
<p>Let the integral part of (8 + <span class="math-tex">\(3 \sqrt{7}\)</span>)<sup>n</sup> = 1, then</p>
<p style="display:inline">I is zero</p>
<p style="display:inline">I is an odd integer</p>
<p style="display:inline">I is an even integer</p>
<p style="display:inline">Nothing can be said about I</p>
Step-by-Step Solution
Key Concept: When (8 + 3√7)^n is expanded using binomial theorem, pairing it with (8 - 3√7)^n yields that their sum is always an integer. Since 0 < (8 - 3√7)^n < 1 for n ≥ 1, the integral part I equals (8 + 3√7)^n + (8 - 3√7)^n minus 1, making I odd.
<p>Let (8 + <span class="math-tex">$3 \sqrt{7}$</span>)<sup>n</sup> = p + f, where p <span class="math-tex">$\in$</span> I and f is a proper fraction.<br />
Let (8 - <span class="math-tex">$3 \sqrt{7}$</span>)<sup>n</sup> = <span class="math-tex">$f^{\prime}$</span>, a proper fraction [<span class="math-tex">$\because$</span> 0 < 8 - <span class="math-tex">$3 \sqrt{7}$</span> < 1]<br />
Since, (8 + <span class="math-tex">$3 \sqrt{7}$</span>)<sup>n</sup> + (8 - <span class="math-tex">$3 \sqrt{7}$</span>)<sup>n</sup> = p + f + <span class="math-tex">$f^{\prime}$</span> is an even integer<br />
<span class="math-tex">$\Leftrightarrow$</span> p + 1 is an even integer ...<br />
[<span class="math-tex">$\because$</span> 0 < f < 1 and 0 < <span class="math-tex">$f^{\prime}$</span> < 1] [<span class="math-tex">$\therefore$</span> 0 < f + <span class="math-tex">$f^{\prime}$</span> < 2 <span class="math-tex">$\Rightarrow$</span> f + <span class="math-tex">$f^{\prime}$</span> = 1]<br />
<span class="math-tex">$\Rightarrow$</span> p is an odd integer.</p>
Correct Answer: B