Value of 2n + 2n-12n+1 - 2n

Option b: 3/2. Solution: We want to simplify the expression: 2n + 2n-12n+1 - 2n Let us factor out the smallest power of 2 from both the numerator and the denominator, which is 2n-1 : - For the numerator: 2n + 2n-1 = 2n-1 × 21 + 2n-1 × 1 = 2n-1(2 + 1) = 3 × 2n-1 - For the denominator: 2n+1 - 2n = 2n-1 × 22 - 2n-1 × 21 = 2n-1(4 - 2) = 2 × 2n-1 Now, substitute these factored forms back into the fraction: 2n + 2n-12n+1 - 2n = 3 × 2n-12 × 2n-1 Since 2n-1 ≠ 0 , we can cancel it from both the numerator and the denominator: = 3/2 This matches Option B. Hence, Option B is the correct answer.

Ratio, Proportion, Indices, LogarithmMCQPYQ May 18Question 873 of 305

All Questions

Value of $\displaystyle \frac{2^n + 2^{n-1}}{2^{n+1} - 2^n}$

Options

\displaystyle \frac{1}{2}

\displaystyle \frac{3}{2}

\displaystyle \frac{2}{3}

\displaystyle \frac{1}{3}

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Correct Answer

✅ Option b — $\displaystyle \frac{3}{2}$

All Options:

A $\displaystyle \frac{1}{2}$
B $\displaystyle \frac{3}{2}$
C $\displaystyle \frac{2}{3}$
D $\displaystyle \frac{1}{3}$

Detailed Solution & Explanation

We want to simplify the expression:

\frac{2^n + 2^{n-1}}{2^{n+1} - 2^n}

Let us factor out the smallest power of

\displaystyle 2

from both the numerator and the denominator, which is

\displaystyle 2^{n-1}

:
- For the numerator:

2^n + 2^{n-1} = 2^{n-1} \cdot 2^1 + 2^{n-1} \cdot 1 = 2^{n-1}(2 + 1) = 3 \cdot 2^{n-1}

- For the denominator:

2^{n+1} - 2^n = 2^{n-1} \cdot 2^2 - 2^{n-1} \cdot 2^1 = 2^{n-1}(4 - 2) = 2 \cdot 2^{n-1}

Now, substitute these factored forms back into the fraction:

\frac{2^n + 2^{n-1}}{2^{n+1} - 2^n} = \frac{3 \cdot 2^{n-1}}{2 \cdot 2^{n-1}}

Since

\displaystyle 2^{n-1} \neq 0

, we can cancel it from both the numerator and the denominator:

= \frac{3}{2}

This matches **Option B**.
Hence, **Option B** is the correct answer.

About This Chapter: Ratio, Proportion, Indices, Logarithm

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Ratio, Proportion, Indices, Logarithms

This chapter covers Ratio, Proportion, Indices, Logarithms and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 5-7 Marks weightage. Focus on understanding core concepts rather than memorizing.

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Value of 2n+2n−12n+1−2n\displaystyle \frac{2^n + 2^{n-1}}{2^{n+1} - 2^n}2n+1−2n2n+2n−1​