Question

Consider the set of numbers $\{1,\,3,\,3^{2},\,3^{3},\,\ldots,\,3^{100}\}$. The ratio of the last number and the sum of the remaining numbers is closest to:

• A. 1
• B. 2
• C. 3
• D. 50
• E. 99

Hint:

For a geometric list with ratio $r>1$, the ratio $\dfrac{\text{last term}}{\text{sum of previous terms}}=\dfrac{(r-1)r^{n-1}}{r^{n-1}-1}\approx r-1$ when $n$ is large. Here $r=3\Rightarrow r-1=2$.

Answer 1

The Correct Option is B

Step 1: Express the required ratio.
Let \[ R=\frac{3^{100}}{1+3+3^{2}+.....s+3^{99}}. \] Step 2: Use the geometric–series sum.
For ratio $3$, the sum of the first $100$ terms excluding the last is \[ S=1+3+3^{2}+.....s+3^{99} =\frac{3^{100}-1}{3-1} =\frac{3^{100}-1}{2}. \] Step 3: Simplify $R$.
\[ R=\frac{3^{100}}{\frac{3^{100}-1}{2}} =\frac{2..... 3^{100}}{3^{100}-1} =\frac{2}{1-3^{-100}}. \] Step 4: Approximate.
Since $3^{-100}$ is extremely small, \[ \frac{1}{1-3^{-100}}\approx 1+3^{-100}\quad\Rightarrow\quad R\approx 2(1+3^{-100})\approx 2\ (\text{just over }2). \] Hence the closest option is $2$. \[ \boxed{2} \]