Let S = 1/25! + 1/(3!23!) + 1/(5!21!) + ... up to 13 terms. If 13S = 2ᵏ/n!, k ∈ N, then n + k is equ

Q: Wait, why do we multiply by $26!$ specifically? How did you know that was the right move?

Great question! In competitive exams like JEE, when you see a sum of terms involving products of factorials in the denominator, your first instinct should be to check the sum of the numbers inside those factorials. If that sum is constant (in this case, $26$), it is a massive signal that the terms are related to binomial coefficients. By multiplying by $26!$, we create the form $\frac{26!}{r!(26-r)!}$, which is exactly $\binom{26}{r}$. It is a standard technique to force the binomial structure.

Q: I forgot the property for the sum of odd binomial coefficients. How do I derive it quickly?

Don't worry, it happens! Remember the binomial expansion of $(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n$. If you set $x=1$, you get the sum of all coefficients as $2^n$. If you set $x=-1$, you get $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \dots = 0$. Subtracting these two equations cancels out the even terms and leaves you with twice the sum of the odd terms, which equals $2^n$. Thus, the sum of odd terms is $2^n / 2 = 2^{n-1}$.

Q: Why are there exactly 13 terms in the series? Does that matter?

It matters immensely! The series is $\frac{1}{25!} + \frac{1}{3!23!} + \dots$. The odd numbers in the denominator are $1, 3, 5, \dots$. Since the last term is $\binom{26}{25}$, we are summing $\binom{26}{1}, \binom{26}{3}, \dots, \binom{26}{25}$. This covers all odd integers from $1$ to $25$. There are exactly $13$ such odd numbers, which confirms that our series is complete and perfectly matches the property $\sum \binom{26}{\text{odd}} = 2^{25}$.

Question

Let $S = \frac{1}{25!} + \frac{1}{3!23!} + \frac{1}{5!21!} + \dots$ up to 13 terms. If $13S = \frac{2^k}{n!}, k \in \mathbb{N}$, then $n + k$ is equal to

Options:

(A) 50

(B) 49

(C) 52

(D) 51

Sigma Percentile · Accepted Answer

### The Beauty of Hidden Symmetry

When you first encounter a series like $S = \frac{1}{25!} + \frac{1}{3!23!} + \frac{1}{5!21!} + \dots$, it is natural to feel a bit overwhelmed. It looks like a chaotic mess of factorials.

But in the world of JEE Advanced, chaos is often just order in disguise. The secret to solving this problem lies not in brute-force calculation, but in pattern recognition. Let us peel back the layers together.

### Phase 1: The Pattern Hunt

Take a deep breath and look at the denominators. In the first term, we have $1!$ and $25!$. Their indices sum to $26$.

In the second term, we have $3!$ and $23!$. Their indices also sum to $26$. As you move to the third term, $5!$ and $21!$, the sum is still $26$.

This is not a coincidence; it is the geometric soul of the problem. Every single term in this series shares this constant sum of $26$. Whenever you see this, your mind should immediately jump to the binomial coefficient formula:

$$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$

### Phase 2: The Binomial Bridge

To turn our series into something manageable, we need to force it into the shape of $\binom{26}{r}$. We have the denominator $r!(26-r)!$, but we are missing the $26!$ in the numerator.

So, let us be bold. We multiply and divide the entire series by $26!$. This gives us:

$$S = \frac{1}{26!} \left[ \frac{26!}{1!25!} + \frac{26!}{3!23!} + \frac{26!}{5!21!} + \dots ight]$$

Suddenly, the chaos vanishes. Each term inside the bracket is now a perfect binomial coefficient: $\binom{26}{1} + \binom{26}{3} + \binom{26}{5} + \dots + \binom{26}{25}$. We have successfully bridged the gap between a scary factorial series and the elegant world of combinations.

### Phase 3: The Summation Magic

Now we are left with the sum of odd binomial coefficients for $n=26$. There is a beautiful property in combinatorics: the sum of all odd-indexed binomial coefficients $\binom{n}{1} + \binom{n}{3} + \dots$ is always $2^{n-1}$.

Since our $n$ is $26$, the sum inside the bracket is simply $2^{26-1}$, which is $2^{25}$. Substituting this back, our series $S$ becomes:

$$S = \frac{2^{25}}{26!}$$

### Phase 4: The Final Polish

We are almost there. The problem asks us to find $13S$. So, we multiply our expression by $13$:

$$13S = 13 	imes \frac{2^{25}}{26!}$$

Do not leave it here! We can simplify $26!$ as $26 	imes 25!$. This allows us to cancel the $13$ with the $26$ in the denominator, leaving a $2$ in the denominator:

$$13S = \frac{13 	imes 2^{25}}{26 	imes 25!} = \frac{1}{2} 	imes \frac{2^{25}}{25!} = \frac{2^{24}}{25!}$$

Comparing this to the target form $\frac{2^k}{n!}$, we immediately see that $k = 24$ and $n = 25$. The final step is to find $n + k$, which is $25 + 24 = 49$.

You have navigated the complexity and arrived at the truth. Remember, in mathematics, the most intimidating problems often have the most elegant solutions. The final answer is **49**.

Animated Solution for Mathematics - Binomial Theorem: Let $S = \frac{1}{25 !} + \frac{1}{3 ! 23 !} + \frac{1}{5 ! 21 !} + \dots$ up to 13 terms. If $13 S = \frac{2 ^{k}}{n !}, k \in N$ , then $n + k$ is equal to

Select Answer:

Visualized Solution

Analyze the Given Series $S$

Observe the Denominator Pattern

Introduce Binomial Coefficients

Rewrite $S$ with $26!$

Convert to Binomial Form

Sum of Odd Binomial Coefficients

Substitute the Sum back into $S$

Calculate $13 S$

Expand $26!$ to Simplify

Cancel Terms and Simplify

Compare and Find $n + k$

The Sigma Insight: Properties of Binomial Coefficients

The Beauty of Hidden Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

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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; } Let S=25!1​+3!23!1​+5!21!1​+… up to 13 terms. If 13S=n!2k​,k∈N, then n+k is equal to

Select Answer:

Visualized Solution

Analyze the Given Series .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Observe the Denominator Pattern

Introduce Binomial Coefficients

Rewrite .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 26!

Convert to Binomial Form

Sum of Odd Binomial Coefficients

Substitute the Sum back into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 13S

Expand .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 26! to Simplify