Binomial Theorem
Grade 11
Question:
<p>If (1 + x)<sup>n</sup> = C<sub>0</sub> + C<sub>1</sub>x + C<sub>2</sub>x<sup>2</sup> + ... + C<sub>n</sub>x<sup>n</sup>, then the value of C<sub>0</sub> + 2C<sub>1</sub> + 3C<sub>2</sub> + ... + (n + 1)C<sub>n</sub> is</p>
<p style="display:inline">(n + 2) 2<sup>n</sup></p>
<p style="display:inline">(n + 1) 2<sup>n-1</sup></p>
<p style="display:inline">(n + 2) 2<sup>n-1</sup></p>
<p style="display:inline">(n + 1) 2<sup>n</sup></p>
Step-by-Step Solution
Key Concept: Split the sum ∑(r+1)Cᵣ into ∑rCᵣ + ∑Cᵣ, then use the identity rCᵣ = n·C_{r-1} to convert the first sum and apply binomial expansions: ∑rCᵣ = n·2^(n-1) and ∑Cᵣ = 2^n.
<p>We have,<br />
C<sub>0</sub> + 2C<sub>1</sub> + 3C<sub>2</sub> + ... + (n + 1) C<sub>n</sub><br />
= <span class="math-tex">$\sum_\limits{r=0}^{n}$</span> (r + 1)C<sub>r</sub><br />
= <span class="math-tex">$\sum_\limits{r=0}^{n}$</span> (r + 1) <sup>n</sup>C<sub>r</sub><br />
= <span class="math-tex">$\sum_\limits{r=0}^{n}$</span> (r<span class="math-tex">$\cdot$</span><sup>n</sup>C<sub>r</sub> + <sup>n</sup>C<sub>r</sub>)<br />
= <span class="math-tex">$\sum_\limits{r=0}^{n}$</span> r<span class="math-tex">$\cdot$</span><sup>n</sup>C<sub>r</sub> + <span class="math-tex">$\sum_\limits{r=0}^{n}$</span> <sup>n</sup>C<sub>r</sub><br />
<span class="math-tex">$=\sum_\limits{r=1}^{n} r \cdot \frac{n}{r}$</span> <sup>n-1</sup>C<sub>r-1</sub> + <span class="math-tex">$\sum_\limits{r=0}^{n}$</span> <sup>n</sup>C<sub>r</sub> ... [<span class="math-tex">$\because$</span> <sup>n</sup>C<sub>r</sub> = <span class="math-tex">$\frac{n}{r}\cdot$</span> <sup>n-1</sup>C<sub>r-1</sub>]<br />
= n(<span class="math-tex">$\sum_\limits{r=1}^{n}$</span> <sup>n-1</sup>C<sub>r-1</sub>) + (<span class="math-tex">$\sum_\limits{r=0}^{n}$</span> <sup>n</sup>C<sub>r</sub>)<br />
= n(<sup>n-1</sup>C<sub>0</sub> + <sup>n-1</sup>C<sub>1</sub> + ... + <sup>n-1</sup>C<sub>n-1</sub>) + (<sup>n</sup>C<sub>0</sub> + <sup>n</sup>C<sub>1</sub> + ... + <sup>n</sup>C<sub>n</sub>)<br />
= n<span class="math-tex">$\cdot$</span>2<sup>n-1</sup> + 2<sup>n</sup><br />
= n<span class="math-tex">$\cdot$</span>2<sup>n-1</sup> + 2<span class="math-tex">$\cdot$</span>(2<sup>n-1</sup>)<br />
= (n + 2)<span class="math-tex">$\cdot$</span>2<sup>n-1</sup></p>
Correct Answer: C