Binomial Theorem
Grade 11
Question:
<p>The value of r for which <span class="math-tex">\(^{20} \mathrm{C}_{r} \ ^{20} \mathrm{C}_{0}+\ ^{20} \mathrm{C}_{r-1}\ ^{20} \mathrm{C}_{1}+\ ^{20} \mathrm{C}_{r-2}\ ^{20} \mathrm{C}_{2}+\ldots .+\ ^{20} \mathrm{C}_{0}\ ^{20} \mathrm{C}_{r}\)</span> is maximum, is</p>
<p style="display:inline">11</p>
<p style="display:inline">10</p>
<p style="display:inline">20</p>
<p style="display:inline">15</p>
Step-by-Step Solution
Key Concept: Identify the given sum as the coefficient of $x^r$ in the product of two binomial expansions using Vandermonde's Identity, then apply the property that the middle term of ${}^nC_r$ is the largest.
<p>We know that, (1 + x)<sup>20</sup> = <sup>20</sup>C<sub>0</sub> + <sup>20</sup>C<sub>1</sub>x<sup>2</sup> + <sup>20</sup>C<sub><span style="font-size: 10.8333px;">2</span></sub>x<sup>2</sup> + ... + <sup>20</sup>C<sub>r-1</sub>x<sup>r-1</sup> + <sup>20</sup>C<sub>r</sub>x<sup>r</sup> + ... + <sup>20</sup>C<sub>20</sub>x<sup>20</sup><br />
<span class="math-tex">$\therefore$</span> (1 + x)<sup>20</sup> <span class="math-tex">$\cdot$</span> (1 + x)<sup>20</sup> = (<sup>20</sup>C<sub>0</sub> + <sup>20</sup>C<sub>1</sub>x + <sup>20</sup>C<sub>2</sub>x<sup>2</sup> + ... + <sup>20</sup>C<sub>r-1</sub> x<sup>r-1</sup> + <sup>20</sup>C<sub>r</sub>x<sup>r</sup> + ... + <sup>20</sup>C<sub>20</sub>x<sup>20</sup>) <span class="math-tex">$\times$</span> (<sup>20</sup>C<sub>0</sub> + <sup>20</sup>C<sub>1</sub>x + ... + <sup>20</sup>C<sub>r-1</sub>x<sup>r-1</sup> + <sup>20</sup>C<sub>r</sub>x<sup>r</sup> + ... + <sup>20</sup>C<sub>20</sub>x<sup>20</sup>)<br />
<span class="math-tex">$\Rightarrow$</span> (1 + x)<sup>40</sup> = (<sup>20</sup>C<sub>0</sub> <span class="math-tex">$\cdot$</span> <sup>20</sup>C<sub>r</sub> + <sup>20</sup>C<sub>1</sub> <sup>20</sup>C<sub>r-1</sub> ... <sup>20</sup>C<sub>r</sub> <sup>20</sup>C<sub>0</sub>) x<sup>r</sup> + ...<br />
On comparing the coefficient of x<sup>r</sup> of both sides, we get<br />
<sup>20</sup>C<sub>0</sub> <sup>20</sup>C<sub>r</sub> + <sup>20</sup>C<sub>1</sub> <sup>20</sup>C<sub>r-1</sub> + ... + <sup>20</sup>C<sub>r</sub> <sup>20</sup>C<sub>0</sub> = <sup>40</sup>C<sub>r</sub><br />
The maximum value of <sup>40</sup>C<sub>r</sub> is possible only when r = 20 [<span class="math-tex">$\because$</span> <sup>n</sup>C<sub>n/2</sub> is maximum when n is even]<br />
Thus, required value of r is 20</p>
Correct Answer: C