Sigma Percentile
JEE Main 2021 (16 March Shift 1)
LEVELBoard

Animated Solution for Mathematics - Binomial Theorem: If is the number of irrational terms in the expansion of , then is divisible by:

Select Answer:

Visualized Solution

Identify the Binomial Expression

  • Given expression:
  • Objective: Find the number of irrational terms
  • Finally, check the divisibility of

The General Term Formula

  • General term formula:
  • In our case: , , and
  • Constraint: and

Substitute Values into General Term

  • Substituting values:

Simplify the Exponents

  • Simplified General Term:

Condition for Rational Terms

  • For rational terms, exponents must be integers:
  • 1.
  • 2.
  • Where

Analyze the Second Constraint

  • From , must be a multiple of
  • Possible values of :

Verify the First Constraint

  • Check for
  • Since is a multiple of and is a multiple of (and thus ), is always a multiple of
  • All values are valid for rational terms.

Count Rational and Total Terms

  • Number of rational terms =
  • Total number of terms in expansion =

Calculate Irrational Terms

  • Number of irrational terms

Find

  • Calculate :

Check Divisibility and Final Answer

  • Check divisibility of :
  • Therefore, is divisible by
  • Correct Option: 26

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

Imagine you are standing before the expression . It looks daunting, but in the world of JEE Advanced, we use the elegance of the Binomial Theorem rather than brute force.
Our mission is to find , the number of irrational terms, and then solve a final divisibility puzzle.

The DNA of the Expansion

Every term in a binomial expansion follows the general term formula:
Here, our , our , and our . Plugging these into the formula, we get:
Using the laws of indices, we simplify this to:

The Rationality Filter

A term is rational if and only if its exponents are integers. If the exponents are fractions, the term involves roots, making it irrational.
We require both and .
Starting with the second condition, implies that must be a multiple of . Given the constraint , our candidates for are:
Now, we check the first condition: . Since is divisible by , we simply need to be divisible by .
Because all our candidates are multiples of , they are automatically multiples of . Thus, all values of produce rational terms.

The Final Count

The total number of terms in the expansion is .
Since there are rational terms, the number of irrational terms is:
The question asks for the value of . Therefore:
The final result is 52.

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