Binomial Theorem
Grade 11

Question:

<p>C<sub>0</sub>&nbsp;- C<sub>1</sub>&nbsp;+ C<sub>2</sub>&nbsp;- C<sub>3</sub>&nbsp;+ ... + (-1)<sup>n</sup>&nbsp;C<sub>n</sub> is equal to</p>
<p style="display:inline">0</p>
<p style="display:inline">2<sup>n</sup></p>
<p style="display:inline">2<sup>n-1</sup></p>
<p style="display:inline">2<sup>n</sup> - 1</p>

Step-by-Step Solution

Key Concept: The alternating sum of binomial coefficients is evaluated by substituting x = -1 into the binomial expansion of (1 + x)^n, which always results in zero.
<p>We know that,<br /> (1 + x)<sup>n</sup>&nbsp;=&nbsp;<sup>n</sup>C<sub>0</sub>&nbsp;+&nbsp;<sup>n</sup>C<sub>1</sub>x +&nbsp;<sup>n</sup>C<sub>2</sub>x<sup>2</sup>&nbsp;+ ... +&nbsp;<sup>n</sup>C<sub>n</sub>x<sup>n</sup><br /> Substituting x = -1, we get<br /> (1 - 1)<sup>n</sup>&nbsp;=&nbsp;<sup>n</sup>C<sub>0</sub>&nbsp;-&nbsp;<sup>n</sup>C<sub>1</sub>&nbsp;+&nbsp;<sup>n</sup>C<sub>2</sub>&nbsp;- ... (-1)<sup>n</sup>&nbsp;<sup>n</sup>C<sub>n</sub><br /> <span class="math-tex">$\therefore$</span>&nbsp;C<sub>0</sub>&nbsp;- C<sub>1</sub>&nbsp;+ C<sub>2</sub>&nbsp;- C<sub>3 </sub>+ ... (-1)<sup>n</sup>&nbsp;C<sub>n</sub> = 0</p>
Correct Answer: A

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