Binomial Theorem
Grade 11
Question:
<p>The coefficient of <span class="math-tex">\(x^{256}\)</span> in the expansion of <span class="math-tex">\((1-x)^{101}\left(x^{2}+x+1\right)^{100}\)</span> is:</p>
<p style="display:inline"><span class="math-tex">\({ }^{100} {C}_{16}\)</span></p>
<p style="display:inline"><span class="math-tex">\(-{ }^{100} {C}_{15}\)</span></p>
<p style="display:inline"><span class="math-tex">\(-{ }^{100} {C}_{16}\)</span></p>
<p style="display:inline"><span class="math-tex">\({ }^{100} C_{15}\)</span></p>
Step-by-Step Solution
Key Concept: Utilize the identity (1-x)(1+x+x^2) = 1-x^3 to transform the product into (1-x)(1-x^3)^100 before applying the binomial theorem.
<p><span class="math-tex">$(1-x)^{101}\left(x^{2}+x+1\right)^{100}$</span><br />
<span class="math-tex">$=(1-x)^{100}\left(x^{2}+x+1\right)^{100}(1-x)$</span><br />
<span class="math-tex">$=\left[(1-x)\left(1+x+x^{2}\right)\right]^{100}(1-x)$</span><br />
<span class="math-tex">$=\left(1-x^{3}\right)^{100}(1-x)$</span><br />
<span class="math-tex">$=(1-x)\left({ }^{100} {C}_{0}-{ }^{100} {C}_{1} x^{3}+{ }^{100} {C}_{2} x^{6}-\ldots \ldots+\right.$</span><span class="math-tex">$\left.{ }^{100} {C}_{84} x^{252}-{ }^{100} {C}_{85} x^{255}+{ }^{100} {C}_{86} x^{258}+\ldots.\right)$</span><br />
<span class="math-tex">$\therefore$</span> Coefficient of <span class="math-tex">$x^{256}$</span> is <span class="math-tex">${ }^{100} {C}_{85}={ }^{100} {C}_{100-85}={ }^{100} {C}_{15}$</span></p>
Correct Answer: D