Find the sum of the series : sum_{r=0}^n (-1)^r nCr [1/2^r + 3^r/2^(2r) + 7^r/2^(3r) + 15^r/2^(4r) .

Q: Why are we allowed to swap the order of summation? Is that always legal?

Great question! In finite summations, as long as the limits are independent (meaning the range of $r$ doesn't depend on $j$ and vice versa), you can always swap them. It is like changing the order of integration in calculus—it is a powerful tool to make the problem solvable. Think of it as rearranging a grid of numbers; the total sum remains the same regardless of whether you add by rows first or by columns first.

Q: I got stuck at the binomial expansion. How did you know $x$ was negative?

Look at the term $(-1)^r$ in the original expression. In the binomial expansion $(1+x)^n = \sum_{r=0}^n {}^nC_r x^r$, the $(-1)^r$ is a dead giveaway that $x$ must be negative. When you group $(-1)^r$ with the fraction $(\frac{2^j-1}{2^j})^r$, you get $(-\frac{2^j-1}{2^j})^r$. That is your $x$! Always look for that negative sign to guide your choice of $x$.

Q: What happens if $m$ goes to infinity? Does the series converge?

That is the spirit of a true mathematician! If $m \to \infty$, the term $\frac{1}{2^{mn}}$ approaches $0$. The sum would then become $\frac{1}{2^n-1}$. It converges beautifully because the common ratio of our G.P. is less than $1$. You have just discovered the infinite sum of this series!

Question

Find the sum of the series : $\sum_{r=0}^n (-1)^r {}^nC_r \left[\frac{1}{2^r} + \frac{3^r}{2^{2r}} + \frac{7^r}{2^{3r}} + \frac{15^r}{2^{4r}} \dots 	ext{ up to } m 	ext{ terms}ight]$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of symbols.

You see a summation, a nested series, powers of $r$, and powers of $2$. It is designed to overwhelm you. But remember, in the world of competitive mathematics, complexity is often just simplicity in disguise.

Let us peel back the layers together.

### The Inner Series

Look closely at the inner part of the expression: 
$$\frac{1}{2^r} + \frac{3^r}{2^{2r}} + \frac{7^r}{2^{3r}} + \dots$$

We need to find the general term, $T_j$. The numerators are $1, 3, 7, 15, \dots$. If you have been practicing your sequences, you know these are not random; they are $2^1-1, 2^2-1, 2^3-1, 2^4-1, \dots$.

The denominators are $2^r, 2^{2r}, 2^{3r}, 2^{4r}, \dots$. When we combine these, the $j$-th term is:
$$T_j = \frac{(2^j-1)^r}{(2^j)^r} = \left(\frac{2^j-1}{2^j}\right)^r$$

We have just turned a scary list of terms into a clean, compact summation:
$$\sum_{j=1}^m \left(\frac{2^j-1}{2^j}\right)^r$$

### The Double Summation

Now, we substitute this back into our original expression $S$. We get:
$$S = \sum_{r=0}^n (-1)^r {}^nC_r \left[ \sum_{j=1}^m \left(\frac{2^j-1}{2^j}\right)^r \right]$$

This is a double summation. In JEE, whenever you see a double sum with independent limits, your first instinct should be: can I swap the order?

By pulling the sum over $j$ to the outside, we get:
$$S = \sum_{j=1}^m \left[ \sum_{r=0}^n {}^nC_r (-1)^r \left(\frac{2^j-1}{2^j}\right)^r \right]$$

Why do this? Because now, the inner sum is a function of $r$ only, and the outer sum is just a spectator.

### The Binomial Collapse

Look at that inner sum: $\sum_{r=0}^n {}^nC_r (-\frac{2^j-1}{2^j})^r$. Does it look familiar?

It is the exact form of the Binomial Theorem:
$$(1+x)^n = \sum_{r=0}^n {}^nC_r x^r$$

Here, our $x$ is $-\frac{2^j-1}{2^j}$. When we apply the theorem, the entire summation collapses into:
$$\left(1 - \frac{2^j-1}{2^j}\right)^n$$

Let us simplify the base:
$$1 - \frac{2^j-1}{2^j} = \frac{2^j - (2^j - 1)}{2^j} = \frac{1}{2^j}$$

So, our monster series has been reduced to:
$$S = \sum_{j=1}^m \left(\frac{1}{2^j}\right)^n = \sum_{j=1}^m \frac{1}{2^{nj}}$$

### The Final Stretch

We are left with a simple Geometric Progression:
$$\frac{1}{2^n} + \frac{1}{2^{2n}} + \frac{1}{2^{3n}} + \dots + \frac{1}{2^{mn}}$$

The first term $a = \frac{1}{2^n}$ and the common ratio $r = \frac{1}{2^n}$. Using the sum formula $S = \frac{a(1-r^m)}{1-r}$, we substitute our values:
$$S = \frac{\frac{1}{2^n}(1 - (\frac{1}{2^n})^m)}{1 - \frac{1}{2^n}}$$

After a bit of algebraic cleanup, the $\frac{1}{2^n}$ terms cancel out, leaving us with the elegant result:
$$S = \frac{2^{mn}-1}{2^{mn}(2^n-1)}$$

You see? The fear was an illusion. By breaking the problem into manageable pieces, identifying the pattern, and using the right tools, we conquered it. Keep this mindset, and no problem will ever be too big for you.

Animated Solution for Mathematics - Binomial Theorem: Find the sum of the series : $\sum_{r = 0}^{n} (- 1)^{r}^{n} C_{r} [\frac{1}{2 ^{r}} + \frac{3 ^{r}}{2 ^{2 r}} + \frac{7 ^{r}}{2 ^{3 r}} + \frac{1 5 ^{r}}{2 ^{4 r}} \dots up to m terms]$ .

Visualized Solution

Analyze the Inner Series

General Term of Inner Series

Rewrite as Double Summation

Swap the Order of Summation

Apply Binomial Theorem

Simplify the Inner Expression

Identify the Geometric Progression

Apply GP Sum Formula

Final Simplification

Conclusion and Key Takeaway

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Inner Series

The Double Summation

The Binomial Collapse

The Final Stretch

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the sum of the series : ∑r=0n​(−1)rnCr​[2r1​+22r3r​+23r7r​+24r15r​… up to m terms].