Binomial Theorem
Grade 11
Question:
<p>Let T<sub>r</sub> be the r<sup>th</sup> term of a sequence, for r = 1, 2, 3, ... If 3T<sub>r+1</sub> = T<sub>r</sub> and T<sub>7</sub> = <span class="math-tex">\(\frac{1}{243}\)</span>, then the value of <span class="math-tex">\(\sum_{r=1}^{\infty}\left(T_{r} \cdot T_{r+1}\right)\)</span> is:</p>
<p style="display:inline"><span class="math-tex">\(\frac{27}{8}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{9}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{81}{4}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{81}{8}\)</span></p>
Step-by-Step Solution
Key Concept: Identify the sequence as a geometric progression and apply the infinite sum formula to the series generated by the products of consecutive terms.
<p><span class="math-tex">\(\frac{\mathrm{T}_{\mathrm{r}+1}}{\mathrm{~T}_{\mathrm{r}}}=\frac{1}{3}\)</span> <br />
If T<sub>1</sub> = a; T<sub>2</sub> = <span class="math-tex">\(\frac{a}{3}\)</span>;<br />
T<sub>3</sub> = <span class="math-tex">\(\frac{\mathrm{a}}{{3}^{2}}\)</span> <br />
<span class="math-tex">\(\therefore\)</span> T<sub>7</sub> = <span class="math-tex">\(\frac{a}{3^{6}}=\frac{1}{243} \Rightarrow \frac{a}{3^{5} \times 3}=\frac{1}{3^{5}}\)</span> <span class="math-tex">\(\Rightarrow\)</span> a = 3<br />
<span class="math-tex">\(\therefore\)</span> T<sub>r</sub> <span class="math-tex">\(\cdot\)</span> T<sub>r+1</sub> = <span class="math-tex">\(\frac{a}{3^{r-1}} \times \frac{a}{3^{r}}=\frac{a^{2}}{3^{2 r-1}}=\frac{3^{2}}{3^{2 r-1}}\)</span> <br />
<span class="math-tex">\(\sum \limits_{{r}=1}^{\infty} {T}_{{r}} \cdot {T}_{{r}+{1}}\)</span> = <span class="math-tex">\(\left(3+\frac{1}{3}+\frac{1}{3^{3}}+\frac{1}{3^{5}}+\ldots \infty\right)\)</span> <br />
= <span class="math-tex">\(\frac{3}{1-\frac{1}{3^{2}}}=\frac{27}{8}\)</span></p>
Correct Answer: A