Definite Integration
Grade 12
Question:
<p>If A = <span class="math-tex">\(\left[\mathbf{a}_{i j}\right]_{n \times n}\)</span>, where a<sub>ij</sub> = i<sup>100</sup> + j<sup>100</sup>, then <span class="math-tex">\(\lim _\limits{n \rightarrow \infty} \frac{\sum_\limits{i=1}^{n} a_{i i}}{n^{101}}\)</span> equals :</p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{101}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{3}{101}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{2}{101}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{50}\)</span></p>
Step-by-Step Solution
Key Concept: Recognize that diagonal elements a_ii = i^100 + i^100 = 2i^100, then convert the sum ∑(i=1 to n) 2i^100 into a Riemann sum approximable by ∫₀¹ 2x^100 dx using the fact that ∑i^100 ≈ n^101∫₀¹ x^100 dx.
<p>We have <span class="math-tex">\(\sum \limits_{i=1}^{n} a_{i i}=a_{11}+a_{22}+\ldots+a_{n n}\)</span><br />
Now, <span class="math-tex">\(\lim \limits_{n \rightarrow \infty} \frac{\sum \limits_{i=1}^{n} a_{i i}}{n^{101}}=\lim \limits_{n \rightarrow \infty} \frac{2\left(1^{100}+2^{100}+\ldots+n^{100}\right)}{n^{101}}\)</span><br />
<span class="math-tex">\(=2 \lim \limits_{n \rightarrow \infty} \frac{1}{n} \sum \limits_{r=1}^{100}\left(\frac{r}{n}\right)^{100}=2 \int_{0}^{1} x^{100} d x=\frac{2}{101}\)</span></p>
Correct Answer: C