Binomial Theorem
Grade None

Question:

<p>If C<sub>0</sub>, C<sub>1</sub>, C<sub>2</sub>, ..., C<sub>n</sub>&nbsp;are the binomial coefficients, then 2<span class="math-tex">\(\cdot\)</span>C<sub>1</sub>&nbsp;+ 2<sup>3</sup><span class="math-tex">\(\cdot\)</span>C<sub>3</sub>&nbsp;+ 2<sup>5</sup><span class="math-tex">\(\cdot\)</span>C<sub>5</sub>&nbsp;+ ... equals</p>
<p style="display:inline"><span class="math-tex">\(\frac{3^{n}+(-1)^{n}}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{3^{n}+1}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{3^{n}-1}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{3^{n}-(-1)^{n}}{2}\)</span></p>

Step-by-Step Solution

Key Concept: Use the difference between the binomial expansions of (1+x)^n and (1-x)^n to isolate terms with odd indices and substitute the specific value of x.
<p>We know that,<br /> [(1 + x)<sup>n</sup>&nbsp;- (1 - x)<sup>n</sup>] = 2[C<sub>1</sub>x + C<sub>3</sub>x<sup>3</sup>&nbsp;+ C<sub>5</sub>x<sup>5</sup>&nbsp;+ ...]<br /> <span class="math-tex">$\Rightarrow \frac{1}{2}$</span>&nbsp;[(1 + x)<sup>n</sup>&nbsp;- (1 - x)<sup>n</sup>] = C<sub>1</sub>x + C<sub>3</sub>x<sup>3</sup>&nbsp;+ C<sub>5</sub>x<sup>5</sup>&nbsp;+ ...<br /> Substituting x = 2, we get,<br /> 2<span class="math-tex">$\cdot$</span>C<sub>1</sub>&nbsp;+ 2<sup>3</sup><span class="math-tex">$\cdot$</span>C<sub>3</sub>&nbsp;+ 2<sup>5</sup><span class="math-tex">$\cdot$</span>C<sub>5</sub>&nbsp;+ ... =&nbsp;<span class="math-tex">$\frac{3^{n}-(-1)^{n}}{2}$</span></p>
Correct Answer: D

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