Definite Integration
Grade None
Question:
<p>A sequence of non-negative terms is given by <span class="math-tex">\(a_{n+1}=\int_\limits{\frac{2}{3}\left(a_{n-1}\right)}^{a_{n}} 3 d x\)</span> where a<sub>0</sub> = 0, a<sub>1</sub> = 1. Then the value (a<sub>1</sub> + a<sub>2</sub> +a<sub>3</sub> + ..... + a<sub>100</sub>) equals :</p>
<p style="display:inline">2<sup>100</sup> - 100</p>
<p style="display:inline">2<sup>101</sup> - 102</p>
<p style="display:inline">2<sup>100</sup></p>
<p style="display:inline">2<sup>100</sup> - 101</p>
Step-by-Step Solution
Key Concept: Convert the integral-based definition into a linear recurrence relation to derive the general term $a_n$ and evaluate the resulting series using the geometric progression sum formula.
<p><span class="math-tex">$a_{n+1}=3\left(a_{n}-\frac{2 a_{n-1}}{3}\right)$</span><br />
a<sub>n+1</sub> = 3a<sub>n</sub> - 2a<sub>n-1</sub><br />
a<sub>2</sub> = 3<br />
a<sub>3</sub> = 7<br />
a<sub>4</sub> = 15 ..... so on<br />
a<sub>n</sub> = 2<sup>n</sup> - 1<br />
<span class="math-tex">$\therefore \quad \sum \limits_{n=1}^{100} a^{n}=\sum \limits_{n=1}^{100}\left(2^{n}-1\right)$</span><span class="math-tex">$=\left(2+2^{2}+\ldots+2^{100}\right)-100$</span><br />
= 2(2<sup>100</sup> - 1) - 100 = 2<sup>101</sup> - 102</p>
Correct Answer: B