Sigma Percentile
JEE Main 2023 (31 January Shift 1)
LEVELBoard

Animated Solution for Mathematics - Binomial Theorem: Let , be the smallest number such that the expansion of has a term . Then is equal to ______.

Enter Numerical Value:

Visualized Solution

Understanding the Binomial Expression

  • Given expression:
  • Target term:
  • Constraints: (smallest possible) and

The General Term Formula

  • General term of :
  • Here,
  • First term
  • Second term

Substituting the Values

  • Substituting into :

Simplifying the First Power of

  • Using the exponent rule:

Simplifying the Second Power of

  • Expanding the second term:

Combining the Exponents of

  • Group constants and terms:
  • Using :
  • Total power of

Comparing with the Target Term

  • The target term is
  • Our term is
  • Comparing the coefficients:
  • Comparing the exponents:

Expressing in terms of

  • From
  • Multiply by :

Applying the Constraint

  • The problem states
  • Substituting our expression for :

Solving the Inequality for

  • Multiply by :
  • Divide by :

Finding the Smallest Integer

  • We know must be an integer ()
  • The inequality is
  • We need the smallest , which corresponds to the smallest valid
  • Smallest integer is

Calculating the Final Value of

  • Substitute into
  • Final Answer:

The Sigma Insight: General Term and Middle Term

Analyzing the Architecture of the Binomial Expansion

We are tasked with analyzing the expression to find a term of the form , where is a natural number and is the smallest positive value possible. This requires a systematic decomposition of the binomial structure.

The General Term

Every binomial expansion follows a specific rhythm defined by the general term formula:
In this structure, our parameters are , , and . Substituting these into the formula, we establish the foundation:

The Algebraic Dance

To simplify the expression, we apply the laws of exponents. First, we process the powers of :
Next, we distribute the power to the second term:
Combining these components by adding the exponents of , we obtain:
Finding a common denominator of , the exponent simplifies to:
Thus, the general term is expressed as:

The Constraint

The problem defines our target term as . By comparing this to our derived expression, we identify the exponent of as :
We are given the constraint that must be the smallest positive number. Therefore, we set the condition :
Solving for :

Final Deduction

Since must be an integer, the smallest integer satisfying is . We now calculate the value of using this integer:
By navigating the constraints and the algebra, we have determined that the smallest positive value for is .

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