Binomial Theorem
Grade 11
Question:
<p>The smallest natural number n, such that the coefficient of x in the expansion of <span class="math-tex">\(\left(x^{2}+\frac{1}{x^{3}}\right)^{n} \text { is }^{n} C_{23}\)</span>, is</p>
<p style="display:inline">35</p>
<p style="display:inline">58</p>
<p style="display:inline">38</p>
<p style="display:inline">23</p>
Step-by-Step Solution
Key Concept: Establish a linear relationship between n and r using the general term formula for the power of x, then apply the binomial identity nCr = nC(n-r) to solve for all possible values of n.
<p>Given binomial is <span class="math-tex">\(\left(x^{2}+\frac{1}{x^{3}}\right)^{n}\)</span>, its (r + 1)<sup>th</sup> term, is <span class="math-tex">\(T_{r+1}=^{n} C_{r}\left(x^{2}\right)^{n-r}\left(\frac{1}{x^{3}}\right)^{r}=^{n} C_{r} x^{2 n-2 r} \frac{1}{x^{3 r}}\)</span> <span class="math-tex">\(=^{n} C_{r} x^{2 n-2 r-3 r}=^{n} C_{r} x^{2 n-5 r}\)</span><br />
For the coefficient of x,<br />
2n - 5r = 1 <span class="math-tex">\(\Rightarrow\)</span>2n = 5r + 1 ...(i)<br />
As coefficient of x is given as <sup>n</sup>C<sub>23</sub>, then either r = 23 or n - r - 23.<br />
If r = 23, then from Eq. (i), we get<br />
2n = 5(23) + 1<br />
<span class="math-tex">\(\Rightarrow\)</span> 2n = 115 + 1 <span class="math-tex">\(\Rightarrow\)</span>2n = 116 <span class="math-tex">\(\Rightarrow\)</span>n = 58<br />
If n - r = 23, then from Eq. (i) on replacing the value of ‘<br />
r', we get 2n = 5(n - 23) + 1<br />
<span class="math-tex">\(\Rightarrow\)</span> 2n = 5n - 115 + 1 <span class="math-tex">\(\Rightarrow\)</span> 3n = 114 <span class="math-tex">\(\Rightarrow\)</span> n = 38<br />
So, the required smallest natural number n = 38</p>
Correct Answer: C