If Cᵣ = ²⁵Cᵣ and C₀ + 5 ⋅ C₁ + 9 ⋅ C₂+...+ (101) ⋅ C₂₅ = 2²⁵ ⋅ k, then k is equal to

Q: Why do we split the summation into two parts? Can't we just solve it directly?

That is a great question! While you could try to expand it, you would be dealing with 26 terms, which is prone to calculation errors. By using the linearity property of summation, we break the problem into two standard, well-known forms. It is not just about making it easier; it is about using the fundamental properties of binomial coefficients to simplify the structure of the problem.

Q: Wait, why does the index of the first summation change from $r=0$ to $r=1$?

Excellent observation! When we use the property $r \binom{n}{r} = n \binom{n-1}{r-1}$, we are essentially dividing by $r$. If $r=0$, the term $r \binom{n}{r}$ is $0 \cdot \binom{n}{0} = 0$. Since the term is zero, it does not contribute to the sum, so we can safely start the index from $r=1$ without changing the total value of the sum.

Q: How do I know when to use the identity $r \binom{n}{r} = n \binom{n-1}{r-1}$?

Whenever you see a binomial coefficient $\binom{n}{r}$ multiplied by a polynomial in $r$ (like $4r+1$), that $r$ is a signal. It is a classic JEE technique to 'absorb' that $r$ into the binomial coefficient using that specific identity. It is your primary tool for reducing the complexity of such series.

Question

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

Sigma Percentile · Accepted Answer

### The Beauty of Binomial Series

My dear student, welcome to a journey through one of the most elegant landscapes in combinatorics. Today, we are not just solving a problem; we are uncovering a hidden structure.

Look at the series: $C_0 + 5C_1 + 9C_2 + \dots + 101C_{25}$. At first glance, it might look like a daunting, long string of numbers. But if you squint, you will see the rhythm.

The multipliers $1, 5, 9, \dots, 101$ are not random. They are dancing to the tune of an Arithmetic Progression with a first term $a=1$ and a common difference $d=4$.

### The Power of Generalization

To master this, we must translate the series into the language of mathematics. The $(r+1)^{	ext{th}}$ term of our A.P. is given by $T_{r+1} = a + rd = 1 + 4r$.

When we pair this with our binomial coefficient $C_r = \binom{25}{r}$, the general term of our series becomes $T_r = (4r + 1) \binom{25}{r}$. Now, we can represent the entire sum $S$ using the compact and powerful sigma notation:

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

Don't let the sigma symbol intimidate you. It is simply a shorthand for 'add everything up.'

### The Art of Decomposition

Here is where the magic happens. We cannot evaluate this sum in one go, so we perform a bit of mathematical surgery. Using the linearity property of summation, we split the sum into two manageable parts:

$$S = 4 \sum_{r=0}^{25} r \binom{25}{r} + \sum_{r=0}^{25} \binom{25}{r}$$

By distributing the binomial coefficient, we have separated the 'complex' part (the one with the $r$) from the 'simple' part (the standard binomial sum).

### The Identity Magic

Let's tackle the second part first. It is a classic identity: the sum of all binomial coefficients $\sum_{r=0}^{n} \binom{n}{r}$ is always $2^n$. With $n=25$, this part is simply $2^{25}$.

Now, for the first part: $4 \sum_{r=0}^{25} r \binom{25}{r}$. That extra $r$ is the trap, but we have the perfect tool to neutralize it: the property $r \binom{n}{r} = n \binom{n-1}{r-1}$. Applying this with $n=25$, our sum becomes:

$$4 \sum_{r=1}^{25} 25 \binom{24}{r-1}$$

Notice how the index shifted to $r=1$? That is because the $r=0$ term is zero. We pull the constant $25$ out, and we are left with $100 \sum_{r=1}^{25} \binom{24}{r-1}$. If we let $k = r-1$, this is just $100 \sum_{k=0}^{24} \binom{24}{k}$, which is $100 \cdot 2^{24}$.

### The Final Synthesis

We are almost there! Our total sum is $S = 100 \cdot 2^{24} + 2^{25}$. To combine these, we make the powers of $2$ identical.

We rewrite $100 \cdot 2^{24}$ as $50 \cdot 2 \cdot 2^{24}$, which is $50 \cdot 2^{25}$. Now, adding them together:

$$S = 50 \cdot 2^{25} + 1 \cdot 2^{25} = (50 + 1) \cdot 2^{25} = 51 \cdot 2^{25}$$

Comparing this to the given form $k \cdot 2^{25}$, we find that **$k = 51$**. You see? With the right tools and a bit of patience, even the most complex series yields to your logic. Keep practicing, and keep falling in love with the process!

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

Enter Numerical Value:

Visualized Solution

Analyzing the Series

General Term of Coefficients

General Term of the Series

Expressing Sum in Sigma Notation

Splitting the Summation

Evaluating the Standard Binomial Sum

Applying the Property $r (r n) = n (r - 1 n - 1)$

Simplifying the First Sum

Evaluating the First Sum

Combining Both Parts

Final Calculation for $k$

The Sigma Insight: Properties of Binomial Coefficients

The Beauty of Binomial Series

The Power of Generalization

The Art of Decomposition

The Identity Magic

The Final Synthesis

Similar Questions

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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