The larger of 99⁵⁰ + 100⁵⁰ and 101⁵⁰ is .........

Q: Why can't we just use logarithms to compare these?

Logarithms are fantastic for products, but they struggle with sums like $99^{50} + 100^{50}$. The logarithm of a sum is not the sum of the logarithms, which makes the algebra messy and unproductive. Stick to the Binomial Theorem; it is the surgical tool designed for this specific structure.

Q: Why do we subtract $99^{50}$ from $101^{50}$?

We are looking for a way to isolate the terms. By bringing $99^{50}$ to the side of $101^{50}$, we create a difference of two binomial expansions. This allows us to exploit the symmetry around $100$, leading to a beautiful cancellation of even-indexed terms.

Q: What happens to the remainder terms?

The remainder terms are all positive. Since we are comparing the difference to $100^{50}$, and we found that the difference is $100^{50}$ plus a strictly positive remainder, it is mathematically certain that the difference is greater than $100^{50}$.

Question

The larger of $99^{50} + 100^{50}$ and $101^{50}$ is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Have you ever stared at an expression like $99^{50} + 100^{50}$ versus $101^{50}$ and felt a wave of intimidation? It is completely natural. These numbers are gargantuan, far beyond the reach of any calculator or standard arithmetic.

In the world of JEE Advanced, we do not solve problems by brute force; we solve them by finding the hidden architecture beneath the surface. Today, we are going to dismantle this giant using the elegance of the **Binomial Theorem**.

### The Symmetry of the Anchor

The first step in any complex problem is to find your anchor. Look at the bases: $99$, $100$, and $101$. They are consecutive integers!

This is not a coincidence; it is a gift. By choosing $100$ as our central anchor, we can rewrite our expression in terms of $100$. We see $99$ as $(100-1)$ and $101$ as $(100+1)$.

Suddenly, the problem transforms from a calculation nightmare into a beautiful algebraic structure:
$$(100+1)^{50} 	ext{ versus } (100-1)^{50} + 100^{50}$$

### The Algebraic Pivot

To make this comparison manageable, let us shift our perspective. Instead of comparing the sum directly, let us move $99^{50}$ to the other side.

Our goal is now to determine if:
$$101^{50} - 99^{50} > 100^{50}$$

This simple rearrangement is the turning point. We are now looking at the difference of two binomial expansions, which is a classic setup for a massive cancellation.

### The Binomial Magnifying Glass

Let us expand both terms using the Binomial Theorem:
$$(x+a)^n = \sum_{r=0}^{n} {^{n}C_{r}} x^{n-r} a^{r}$$

For $(100+1)^{50}$, every term is positive. For $(100-1)^{50}$, the signs alternate because of the $(-1)^r$ factor.

When we subtract $(100-1)^{50}$ from $(100+1)^{50}$, the even-indexed terms (where $r$ is $0, 2, 4, \dots$) have the same sign in both expansions and thus cancel out perfectly. The odd-indexed terms (where $r$ is $1, 3, 5, \dots$) have opposite signs, so when we subtract, they double up.

The result is:
$$2 	imes \left[ {^{50}C_1} 100^{49} + {^{50}C_3} 100^{47} + \dots + {^{50}C_{49}} 100^1 ight]$$

### The Final Calculation

Now, look at the very first term of this series: $2 	imes {^{50}C_1} 	imes 100^{49}$. Since ${^{50}C_1} = 50$, this becomes:
$$2 	imes 50 	imes 100^{49} = 100 	imes 100^{49} = 100^{50}$$

This is the magic moment! We have shown that the difference $101^{50} - 99^{50}$ is exactly $100^{50}$ plus a remaining series of positive terms.

Because that remainder is strictly positive, the difference must be strictly greater than $100^{50}$. Therefore, **$101^{50} > 99^{50} + 100^{50}$**.

Animated Solution for Mathematics - Binomial Theorem: The larger of $9 9^{50} + 10 0^{50}$ and $10 1^{50}$ is .........

Visualized Solution

The Challenge of Large Powers

Shifting the Perspective

Expressing via the Central Anchor

Recalling the Binomial Theorem

The Magic of Cancellation

Analyzing the Leading Term

The Power of the Remainder

The Final Verdict

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

The Symmetry of the Anchor

The Algebraic Pivot

The Binomial Magnifying Glass

The 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; } The larger of 9950+10050 and 10150 is .........