Question

Let \(X\) be a random variable such that
\[ P(X=i)=2P(X=i-1), \quad i=2,3,\ldots,n,\ n\geq7, \]
and \(\displaystyle\sum_{i=1}^{n}P(X=i)=1\). Then \((2^n-1)P(X=7)\) equals ________ (answer in integer).

Hint:

Hint:
The recursion \(P(X=i)=2P(X=i-1)\) makes \(\{P(X=i)\}\) a geometric sequence. Write \(P(X=i)=2^{i-1}P(X=1)\) and use the normalization condition to find \(P(X=1)\).

Answer 1

Correct Answer: 64

Step 1: Read the recursion.
The rule \(P(X=i)=2P(X=i-1)\) for \(i=2,3,\ldots,n\) says each probability is exactly double the one before it. This is a geometric progression with common ratio 2, starting from \(P(X=1)\).

Step 2: Write a general formula for P(X=i).
Applying the recursion repeatedly,
\[ P(X=2)=2P(X=1),\quad P(X=3)=2P(X=2)=2^2P(X=1),\ \ldots \]
so in general
\[ P(X=i)=2^{i-1}P(X=1), \qquad i=1,2,\ldots,n. \]

Step 3: Apply the normalization condition.
Since the probabilities must sum to 1,
\[ \sum_{i=1}^{n}P(X=i)=P(X=1)\sum_{i=1}^{n}2^{i-1}=1. \]

Step 4: Sum the geometric series.
The sum of a geometric series with first term 1 and ratio 2, taken over n terms, is
\[ \sum_{i=1}^{n}2^{i-1}=\frac{2^n-1}{2-1}=2^n-1. \]
So
\[ P(X=1)(2^n-1)=1 \quad \Rightarrow \quad P(X=1)=\frac{1}{2^n-1}. \]

Step 5: Find P(X=7).
Using the general formula from Step 2 with \(i=7\),
\[ P(X=7)=2^{6}P(X=1)=\frac{64}{2^n-1}. \]

Step 6: Compute the required quantity.
\[ (2^n-1)P(X=7)=(2^n-1)\cdot\frac{64}{2^n-1}=64. \]

Final Answer:
The value of \((2^n-1)P(X=7)\) does not depend on n and equals 64. \[ \boxed{64} \]