Sigma Percentile
JEE Advanced 2013
LEVELBoard

Animated Solution for Mathematics - Binomial Theorem: The coefficients of three consecutive terms of are in the ratio . Then

Enter Numerical Value:

Visualized Solution

Identifying the Terms

  • Expansion of
  • Let the three consecutive terms be , , and .
  • Their coefficients are , , and .

The Binomial Coefficient Formula

  • To avoid expanding factorials, we use the ratio formula:
  • In our case, the index .

Setting up the First Ratio

  • Ratio of the first two coefficients is .

Applying Formula to First Ratio

  • Invert the ratio to match the formula structure:
  • Substitute and into the formula:

Simplifying the First Equation

  • Simplify the numerator:
  • Cross-multiply:
  • Rearrange to get Equation 1:

Setting up the Second Ratio

  • Ratio of the next two coefficients is .

Applying Formula to Second Ratio

  • Invert the ratio:
  • Substitute and into the formula:

Simplifying the Second Equation

  • Simplify the numerator:
  • Cross-multiply:
  • Expand and rearrange to get Equation 2:

Solving the System of Equations

  • We have a system of two linear equations:
  • 1)
  • 2)
  • Multiply Equation 1 by to match the term:

Finding the Value of

  • Equate the two expressions for :
  • Rearrange to solve for :
  • Final Answer:

The Sigma Insight: Properties of Binomial Coefficients

Solution Diagram

Analyzing the Setup

The expansion of involves binomial coefficients that follow the structure of Pascal's Triangle. We are given three consecutive coefficients in the ratio .
To solve this efficiently, we avoid the cumbersome factorial definition . Instead, we utilize the Ratio Property of binomial coefficients.
For any two consecutive binomial coefficients, the ratio is defined as:
This property serves as the master key to unlocking the relationship between the terms without expanding large factorials.

Phase 1

The First Link
Let the three consecutive coefficients be , , and . The ratio of the first two is , which simplifies to .
We set up the ratio as:
Applying the ratio property with and :
Simplifying this expression leads to our first anchor equation:

Phase 2

The Second Link
Next, we consider the second ratio, , which simplifies to . We set up the ratio of the second and third terms:
Here, our is . Substituting this into the ratio property:
Cross-multiplying yields , which expands to . Rearranging the terms gives us our second equation:

The Final Convergence

We now possess a system of two linear equations:
1)
2)
To solve, we multiply the first equation by to align the terms:
Equating the two expressions for :
Solving for , we find the final result:

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