Binomial Theorem
Grade 11
Question:
<p>If x = 1 + <span class="math-tex">\(\frac{3}{1 !} \times \frac{1}{6}+\frac{3 \times 7}{2 !}\left(\frac{1}{6}\right)^{2}+\frac{3 \times 7 \times 11}{3 !}\left(\frac{1}{6}\right)^{3}\)</span> + ... then x<sup>4</sup> =</p>
<p style="display:inline">8</p>
<p style="display:inline">27</p>
<p style="display:inline">81</p>
<p style="display:inline">54</p>
Step-by-Step Solution
Key Concept: Recognize the series as a generalized binomial expansion of the form (1+a)^n where the numerators follow the pattern 3, 3×7, 3×7×11,... representing the Pochhammer symbol (3)_r. The series sums to (1 + 1/6)^3 = (7/6)^3, then compute x^4 = ((7/6)^3)^4 = (7/6)^12.
<p>x = 1 + <span class="math-tex">$\frac{3}{1 !} \times \frac{1}{6}+\frac{3 \times 7}{2 !}\left(\frac{1}{6}\right)^{2}+\frac{3 \times 7 \times 11}{3 !}\left(\frac{1}{6}\right)^{3}$</span> + ...<br />
<span class="math-tex">$\Leftrightarrow$</span> x = <span class="math-tex">$\left(1-\frac{2}{3}\right)^{\frac{-3}{4}} \Leftrightarrow x=\left(\frac{1}{3}\right)^{\frac{-3}{4}}$</span><br />
<span class="math-tex">$\Leftrightarrow$</span> x = <span class="math-tex">$3^{\frac{3}{4}}\Rightarrow$</span> x<sup>4</sup> = 3<sup>3</sup> = 27</p>
Correct Answer: B