If Cr = 25Cr and C0 + 5 · C1 + 9 · C2 + ... + 101 · C25 = 2^25 · k then k is equal to.

Q: Wait, why do we reverse the series? Is there a formula for this?

While there are general formulas for series involving an A.P. multiplied by binomial coefficients, the 'reversal method' is a powerful, intuitive technique that works because of the inherent symmetry of binomial coefficients. It transforms a complex-looking sum into a much simpler one by exploiting the fact that $^{n}C_r = ^{n}C_{n-r}$. It is a classic JEE trick that saves time and reduces the chance of algebraic errors.

Q: Why does the sum of the coefficients become constant when we add the series to its reverse?

This is the beauty of the symmetry property! When you add the $r$-th term of the original series, $(4r+1)C_r$, to the $r$-th term of the reversed series, $(101-4r)C_r$, the variable $r$ cancels out: $(4r+1) + (101-4r) = 102$. Since this sum is independent of $r$, it remains constant for every term in the series, allowing us to factor it out.

Q: Do I really need to calculate $2^{25}$ to find $k$?

Absolutely not! The problem asks for $k$ in the expression $S = 2^{25} \cdot k$. By simplifying the sum, you will find that $S$ is a multiple of $2^{25}$. The $2^{25}$ term will naturally appear in your calculation and will cancel out when you equate it to $k \cdot 2^{25}$, leaving you with the value of $k$ directly.

Question

If $C_r = ^{25}C_r$ and $C_0 + 5 \cdot C_1 + 9 \cdot C_2 + \dots + 101 \cdot C_{25} = 2^{25} \cdot k$ then $k$ is equal to__________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE Advanced mastery. Today, we are going to dissect a problem that, at first glance, might seem like a tedious exercise in arithmetic.

We are looking at the series $S = C_0 + 5 \cdot C_1 + 9 \cdot C_2 + \dots + 101 \cdot C_{25}$. Our goal is to find the constant $k$ such that $S = 2^{25} \cdot k$.

### The Hidden Pattern

Let us pause and observe the structure of this series. We have a sum where each term is a product of a number and a binomial coefficient.

The numbers multiplying these coefficients are $1, 5, 9, \dots, 101$. These form an Arithmetic Progression (A.P.) with a first term $a = 1$ and a common difference $d = 4$.

The general term of this A.P. is $4r + 1$. Thus, our series can be written as:

$$S = \sum_{r=0}^{25} (4r+1) \cdot ^{25}C_r$$

This structure is a classic signature of a problem that can be solved using the symmetry of binomial coefficients.

### The Mirror Technique

In the world of competitive mathematics, when you see a series involving binomial coefficients, always look for symmetry. The most powerful tool in our arsenal is the property $^{n}C_r = ^{n}C_{n-r}$.

This property tells us that the coefficients are symmetric from both ends. For $n = 25$, this means $C_0 = C_{25}$, $C_1 = C_{24}$, and so on.

Now, let us apply the 'reversal method'. We write the series $S$ in reverse order:

$$S = 101 \cdot C_{25} + 97 \cdot C_{24} + 93 \cdot C_{23} + \dots + 1 \cdot C_0$$

We haven't changed the value of the sum; we have simply changed the order of addition. Now, let us apply our symmetry property to this reversed series. Replacing $C_{25}$ with $C_0$, $C_{24}$ with $C_1$, and so on, the reversed series transforms into:

$$S = 101 \cdot C_0 + 97 \cdot C_1 + 93 \cdot C_2 + \dots + 1 \cdot C_{25}$$

### The Magic Addition

This is where the magic happens. Let us add our original series and this newly transformed reversed series vertically, term by term. On the left side, $S + S$ gives us $2S$.

On the right side, for each term $C_r$, we add the coefficients:

$$2S = (1 + 101)C_0 + (5 + 97)C_1 + (9 + 93)C_2 + \dots + (101 + 1)C_{25}$$

Look at the sum of the coefficients: $1 + 101 = 102$, $5 + 97 = 102$, $9 + 93 = 102$. The sum is constant! Every single term now has a common factor of $102$.

We can factor it out:

$$2S = 102 \cdot (C_0 + C_1 + C_2 + \dots + C_{25})$$

### Final Calculation

We are now left with the sum of all binomial coefficients from $C_0$ to $C_{25}$. Recall the fundamental identity from the binomial theorem:

$$\sum_{r=0}^{n} {^{n}C_r} = 2^n$$

For $n = 25$, this sum is exactly $2^{25}$. Substituting this back into our equation, we get:

$$2S = 102 \cdot 2^{25}$$

To find $S$, we simply divide both sides by $2$:

$$S = \frac{102}{2} \cdot 2^{25} = 51 \cdot 2^{25}$$

Comparing this with the given expression $S = k \cdot 2^{25}$, it is crystal clear that **$k = 51$**.

Animated Solution for Mathematics - Binomial Theorem: If $C_{r} =^{25} C_{r}$ and $C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + \dots + 101 \cdot C_{25} = 2^{25} \cdot k$ then $k$ is equal to__________.

Enter Numerical Value:

Visualized Solution

Analyzing the Given Series

The Symmetry of Binomial Coefficients

Writing the Reversed Series

Substituting Symmetrical Coefficients

Adding the Original and Reversed Series

Simplifying the Sum

Sum of All Binomial Coefficients

Finding the Value of $S$

Final Comparison to Find $k$

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Hidden Pattern

The Mirror Technique

The Magic Addition

Final Calculation

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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; } If Cr​=25Cr​ and C0​+5⋅C1​+9⋅C2​+⋯+101⋅C25​=225⋅k then k is equal to__________.