Let m, n ∈ N and gcd (2, n) = 1. If 30 (30 over 0) + 29 (30 over 1) + ....+2 (30 over 28) +1 (30 ove

Q: Wait, why can we just add the $r=30$ term? Doesn't that change the sum?

That is a fantastic observation! Adding the $r=30$ term means adding $(30-30) \binom{30}{30}$, which is $0 \cdot 1 = 0$. Since we are adding zero, the value of the series remains unchanged. We do this to complete the symmetry of the summation, which allows us to use the standard binomial identities from $r=0$ to $n$ without worrying about missing the final term.

Q: How do I know which identity to use for $\sum r \binom{n}{r}$?

Whenever you see a summation where the variable $r$ is multiplied by a binomial coefficient $\binom{n}{r}$, it is a massive red flag that you need the identity $\sum_{r=0}^{n} r \binom{n}{r} = n 2^{n-1}$. This identity is derived from differentiating the binomial expansion of $(1+x)^n$. Whenever you see that $r$ hanging out in front, think of this identity immediately!

Q: Why do we need to extract a factor of 2 from 30 at the end?

The problem explicitly states $\gcd(2, n) = 1$. This is a formal mathematical way of saying '$n$ must be an odd integer'. Since our initial calculation gave us $30 \cdot 2^{29}$, and $30$ is an even number, we cannot leave it like that. We must factor out the $2$ from $30$ (writing it as $15 \cdot 2$) to satisfy the condition that $n$ is odd.

Question

Let $m, n \in N$ and gcd $(2, n) = 1$. If $30 \binom{30}{0} + 29 \binom{30}{1} + ....+2 \binom{30}{28} +1 \binom{30}{29} = n.2^m$, then $n + m$ is equal to

Sigma Percentile · Accepted Answer

### The Art of Pattern Recognition: Unlocking the Binomial Series

Have you ever stared at a long, intimidating series and felt like you were looking at a wall of numbers? I know that feeling.

In JEE Advanced, problems like this are not tests of your ability to calculate massive numbers; they are tests of your ability to see the hidden structure beneath the chaos. Let's break down this series together and find the elegance hidden within.

### Phase 1: The Hidden Symmetry

We are given the series $S = 30 \binom{30}{0} + 29 \binom{30}{1} + \dots + 1 \binom{30}{29}$. At first glance, it looks like a mess. But look closer at the multiplier and the lower index of the binomial coefficient in each term.

For the first term, we have $30$ and $0$. Their sum is $30$. For the second term, we have $29$ and $1$. Their sum is $30$. For the last term, we have $1$ and $29$. Their sum is $30$.

Do you see it? The sum of the multiplier and the index is constant! This is the 'Aha!' moment. We can define the general term $T_r$ as $(30 - r) \binom{30}{r}$. This realization transforms a confusing list of numbers into a beautiful, compact mathematical expression:

$$S = \sum_{r=0}^{29} (30 - r) \binom{30}{r}$$

### Phase 2: The Power of Zero

Now, we have a summation, but the upper limit is $29$. Our standard binomial identities usually work from $r=0$ to $n$. Can we extend this?

Let's check the term for $r=30$. It would be $(30 - 30) \binom{30}{30} = 0 \cdot 1 = 0$.

Since adding zero doesn't change the value of our sum, we can safely extend the upper limit to $30$. Now our series is $S = \sum_{r=0}^{30} (30 - r) \binom{30}{r}$. This symmetry is exactly what we need to unleash our binomial toolkit.

### Phase 3: Linearity and the Toolkit

Let's distribute the terms inside the summation. We get $S = \sum_{r=0}^{30} \left[ 30 \binom{30}{r} - r \binom{30}{r} \right]$. Using the linearity property of summations, we can split this into two distinct parts:

$$S = 30 \sum_{r=0}^{30} \binom{30}{r} - \sum_{r=0}^{30} r \binom{30}{r}$$

Now, we call upon two of the most powerful identities in your JEE arsenal. First, the sum of all binomial coefficients: $\sum_{r=0}^{n} \binom{n}{r} = 2^n$. Second, the sum involving the index: $\sum_{r=0}^{n} r \binom{n}{r} = n \cdot 2^{n-1}$.

These are not just formulas; they are shortcuts through the forest of algebra. For $n=30$, these become:

1. $\sum_{r=0}^{30} \binom{30}{r} = 2^{30}$
2. $\sum_{r=0}^{30} r \binom{30}{r} = 30 \cdot 2^{29}$

### Phase 4: The Final Simplification

Substituting these back into our expression for $S$, we get:

$$S = 30 \cdot (2^{30}) - (30 \cdot 2^{29})$$

Don't reach for a calculator! We can factor out the common term $30 \cdot 2^{29}$:

$$S = 30 \cdot 2^{29} (2^1 - 1)$$

Since $2^1 - 1 = 1$, the expression collapses beautifully to $S = 30 \cdot 2^{29}$.

### Phase 5: The Final Trap

We are almost there, but wait! The problem states $S = n \cdot 2^m$ with the condition $\gcd(2, n) = 1$. This condition is a subtle trap. It tells us that $n$ must be an odd integer.

Currently, our $n$ is $30$, which is even. We must extract a factor of $2$ from $30$ to satisfy the condition:

$$30 = 15 \cdot 2$$

So, $S = (15 \cdot 2) \cdot 2^{29} = 15 \cdot 2^{30}$.

Comparing this to $n \cdot 2^m$, we identify $n = 15$ and $m = 30$. Since $15$ is odd, the condition $\gcd(2, 15) = 1$ is satisfied. Finally, the question asks for $n + m$, which is **$15 + 30 = 45$**.

Animated Solution for Mathematics - Binomial Theorem: Let $m, n \in N$ and gcd $(2, n) = 1$ . If $30 (0 30) + 29 (1 30) + .... + 2 (28 30) + 1 (29 30) = n . 2^{m}$ , then $n + m$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Given Series $S$

Identify the General Term $T_{r}$

Express $S$ as a Summation

Split the Summation

Standard Binomial Identities

Apply Identities for $n = 30$

Simplify the Expression

Match with $n \cdot 2^{m}$

Determine $n$ and $m$

Calculate $n + m$

The Sigma Insight: Properties of Binomial Coefficients

The Art of Pattern Recognition

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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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 m,n∈N and gcd (2,n)=1. If 30(030​)+29(130​)+....+2(2830​)+1(2930​)=n.2m, then n+m is equal to

Enter Numerical Value:

Visualized Solution

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Identify the General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tr​

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Split the Summation

Standard Binomial Identities

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Simplify the Expression

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