If ¹¹C₁/2 + ¹¹C₂/3 + ... + ¹¹C₉/10 = n/m with gcd(n, m) = 1, then n + m is equal to

Q: Wait, why do we reject the negative root here? Or rather, why can't we just use energy conservation?

This is a common confusion! In this problem, we are dealing with a summation of binomial coefficients, not a physics problem involving energy. The 'trap' here is trying to sum the terms individually, which is tedious and prone to error. The key is to use the identity $\frac{^nC_r}{r+1} = \frac{^{n+1}C_{r+1}}{n+1}$ to simplify the expression into a manageable form.

Q: Why do we subtract $^{12}C_0, ^{12}C_1, ^{12}C_{11},$ and $^{12}C_{12}$?

Great question! The total sum of binomial coefficients for $n=12$ is $\sum_{k=0}^{12} ^{12}C_k = 2^{12}$. Our series, after transformation, is $\sum_{r=1}^9 ^{12}C_{r+1}$, which expands to $^{12}C_2 + ^{12}C_3 + \dots + ^{12}C_{10}$. Since our sum only includes terms from $k=2$ to $k=10$, we must subtract the terms for $k=0, 1, 11,$ and $12$ from the total sum.

Q: How do we know $\gcd(2035, 6) = 1$?

To check if $\gcd(2035, 6) = 1$, we check if $2035$ is divisible by the prime factors of $6$, which are $2$ and $3$. $2035$ is odd, so it is not divisible by $2$. The sum of the digits of $2035$ is $2+0+3+5 = 10$, which is not divisible by $3$. Since it is not divisible by $2$ or $3$, it is not divisible by $6$. Thus, they are coprime.

Question

If $\frac{^{11}C_1}{2} + \frac{^{11}C_2}{3} + \dots + \frac{^{11}C_9}{10} = \frac{n}{m}$ with $\gcd(n, m) = 1$, then $n + m$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with evaluating the sum:
$$S = \frac{^{11}C_1}{2} + \frac{^{11}C_2}{3} + \dots + \frac{^{11}C_9}{10}$$

If you attempt to calculate each term individually, you will quickly find yourself lost in a forest of arithmetic. Mathematics is the art of finding the elegant path through the chaos.

### The Hidden Identity

The first thing that should catch your eye is the structure of the general term:
$$T_r = \frac{^{11}C_r}{r+1}$$

The variable $r$ in the denominator prevents us from using standard summation formulas directly. We need a "magic spell" to absorb that $r$.

That spell is the powerful binomial identity:
$$\frac{^nC_r}{r+1} = \frac{^{n+1}C_{r+1}}{n+1}$$

By applying this, we transform the variable denominator into a constant. For our case, where $n=11$, the term becomes:
$$\frac{^{12}C_{r+1}}{12}$$

Suddenly, the denominator is just a constant $12$, which we can pull out of the summation entirely.

### The Missing Pieces

Now, our sum looks like this:
$$S = \frac{1}{12} \sum_{r=1}^9 {^{12}C_{r+1}}$$

Expanding this, we get:
$$S = \frac{1}{12} [^{12}C_2 + ^{12}C_3 + \dots + ^{12}C_{10}]$$

We know that the sum of all coefficients $\sum_{k=0}^{12} {^{12}C_k}$ is $2^{12} = 4096$. However, our bracket is missing the terms for $k=0, 1, 11,$ and $12$.

We must subtract these "missing pieces" from our total. Calculating them is straightforward:
$$^{12}C_0 = 1, \quad ^{12}C_1 = 12, \quad ^{12}C_{11} = 12, \quad ^{12}C_{12} = 1$$

The sum of these missing terms is $1 + 12 + 12 + 1 = 26$.

### The Final Tally

Subtracting these from our total, we get $4096 - 26 = 4070$. Now, we bring back our constant factor:
$$S = \frac{4070}{12}$$

Simplifying this fraction by dividing both numerator and denominator by $2$, we arrive at:
$$S = \frac{2035}{6}$$

Since $2035$ and $6$ share no common factors, we have found our values. The final step is to add them:
**$n + m = 2035 + 6 = 2041$**

And there you have it! We have navigated the complexity and arrived at the solution. Remember, in JEE Advanced, it is rarely about brute force; it is about recognizing the underlying structure and applying the right tool.

Animated Solution for Mathematics - Binomial Theorem: If $\frac{^{11} C _{1}}{2} + \frac{^{11} C _{2}}{3} + \dots + \frac{^{11} C _{9}}{10} = \frac{n}{m}$ with $g cd (n, m) = 1$ , then $n + m$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Given Series

Identify the Binomial Identity

Apply Identity to the General Term

Rewrite the Summation

Recall the Sum of Coefficients

Identify Missing Terms

Calculate Values of Missing Terms

Subtract to Find the Bracket Sum

Finalize the Fraction n/m

Calculate Final Answer n + m

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Hidden Identity

The Missing Pieces

The Final Tally

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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 211C1​​+311C2​​+⋯+1011C9​​=mn​ with gcd(n,m)=1, then n+m is equal to