Sum of the first n terms of the series 1/2 + 3/4 + 7/8 + 15/16 + ......... is equal to (A) 2ⁿ - n -

Q: Wait, why can't we just use the standard formula for an Arithmetic or Geometric Progression directly on the original series?

That is a fantastic question! If you look closely at the terms $\frac{1}{2}, \frac{3}{4}, \frac{7}{8}$, you will notice that the difference between consecutive terms is not constant (so it's not an AP), and the ratio between consecutive terms is not constant (so it's not a GP). It is a hybrid series. We must decompose it into simpler, recognizable parts first.

Q: How did you know to subtract each term from $1$? Is there a general rule for this?

In JEE problems, when you see terms like $\frac{1}{2}, \frac{3}{4}, \frac{7}{8}$, they are 'almost' integers. Whenever you see a series approaching a limit (like $1$), always check the difference between the term and that limit. It is a classic pattern-recognition technique called the 'method of differences' or 'complementary series'.

Q: Why do we treat the sum of $1$ as $n$? Is that always true?

Yes, absolutely! When you sum the constant $1$ from $r=1$ to $n$, you are essentially adding $1 + 1 + 1 + \dots$ ($n$ times). Mathematically, $\sum_{r=1}^{n} 1 = n \times 1 = n$. It is a fundamental property of summation that often gets overlooked in the heat of complex calculations.

Question

Sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is equal to

Options:

(A) $2^n - n - 1$

(B) $1 - 2^{-n}$

(C) $n + 2^{-n} - 1$

(D) $2^n + 1$

Sigma Percentile · Accepted Answer

### The Art of Seeing the Invisible

Welcome, future engineers! Today, we are going to peel back the layers of a series that looks intimidating but hides a beautiful simplicity. We are looking at the series:

$$ \frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots $$

At first glance, it seems like a chaotic mess of fractions, but in the world of JEE Advanced, chaos is just order waiting to be discovered. Let us dive in.

### The Observation

Look at the terms. $\frac{1}{2}$ is $0.5$, $\frac{3}{4}$ is $0.75$, and $\frac{7}{8}$ is $0.875$. Do you see it? They are all dancing around the number $1$.

They are "almost" integers. This is our key. Instead of trying to sum these fractions directly, which would be a nightmare of common denominators, let us look at what they are missing.

For the first term, the gap to reach $1$ is $\frac{1}{2}$. For the second, it is $\frac{1}{4}$. For the third, it is $\frac{1}{8}$. The gaps are powers of $\frac{1}{2}$!

### The General Term

This realization allows us to rewrite every single term in a much smarter way. We can define the general term $T_r$ at position $r$ as:

$$ T_r = 1 - \frac{1}{2^r} $$

This is the "Aha!" moment. By transforming the series into this form, we have moved from a complex, unrecognizable sequence to a structure we can easily manipulate.

### The Summation

Our goal is to find the sum of the first $n$ terms, $S_n = \sum_{r=1}^{n} T_r$. Substituting our new expression, we get:

$$ S_n = \sum_{r=1}^{n} \left( 1 - \frac{1}{2^r} \right) $$

By the linearity of summation, we can split this into two separate, manageable sums:

$$ S_n = \sum_{r=1}^{n} 1 - \sum_{r=1}^{n} \frac{1}{2^r} $$

### The Geometric Progression

The first part is trivial: summing $1$, $n$ times, simply gives us $n$. The second part, $\sum_{r=1}^{n} \frac{1}{2^r}$, is a classic Geometric Progression (GP) where the first term $a = \frac{1}{2}$ and the common ratio $r = \frac{1}{2}$.

Using the trusty GP sum formula $S_{GP} = \frac{a(1-r^n)}{1-r}$, we substitute our values:

$$ S_{GP} = \frac{\frac{1}{2}(1 - (\frac{1}{2})^n)}{1 - \frac{1}{2}} $$

The denominator $1 - \frac{1}{2}$ is $\frac{1}{2}$, which cancels perfectly with the numerator, leaving us with $1 - (\frac{1}{2})^n$, or $1 - 2^{-n}$.

### The Grand Finale

Putting it all together, we return to our main equation:

$$ S_n = n - (1 - 2^{-n}) $$

Opening the bracket, we get the final result:

$$ **S_n = n - 1 + 2^{-n}** $$

And there you have it! The elegance of mathematics revealed. The key takeaway is to always look for hidden patterns that can simplify your terms. You have just conquered a series that would have stumped the unprepared. Keep that curiosity alive!

Animated Solution for Mathematics - Sequence and Series: Sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is equal to

Select Answer:

Visualized Solution

Observe the Pattern

Analyze the Gaps

General Term $T_{r}$

Set up the Summation $S_{n}$

Regrouping Terms

Sum of Constants

Identify Geometric Progression

Sum of GP Formula

Substitute into GP Formula

Simplify the Denominator

Cancel and Simplify $S_{GP}$

Final Calculation for $S_{n}$

Conclusion

The Sigma Insight: Sum of Special Series

The Art of Seeing the Invisible

The Observation

The General Term

The Summation

The Geometric Progression

The Grand Finale

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