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Maths Question 6 – JEE-MAIN 2026

If for 3r30, (3030r)+3(3031r)+3(3032r)+(3033r)=(mr), then m equals:

Recall the symmetry property of binomial coefficients, (nk)=(nnk), to simplify the terms.

Step 1: Apply Symmetry Property✦ Active

First, use the property (nk)=(nnk) to rewrite the given terms. This simplifies the lower index of the binomial coefficients.

(3030r)=(30r) (3031r)=(3030(31r))=(30r1) (3032r)=(3030(32r))=(30r2) (3033r)=(3030(33r))=(30r3)

The expression becomes:

(30r)+3(30r1)+3(30r2)+(30r3)
Step 2: Apply Pascal's Identity Iteratively○ Expand

Rearrange the terms and apply Pascal's identity, (nk)+(nk1)=(n+1k), multiple times. We can write 3(30r1) as (30r1)+2(30r1) and 3(30r2) as 2(30r2)+(30r2).

((30r)+(30r1))+2((30r1)+(30r2))+((30r2)+(30r3))

Applying Pascal's identity to each pair:

(31r)+2(31r1)+(31r2)

Now, split 2(31r1) into (31r1)+(31r1) and apply Pascal's identity again:

((31r)+(31r1))+((31r1)+(31r2))

Applying Pascal's identity:

(32r)+(32r1)
💡 Teacher's Secret Hint

Ensure the lower index for Pascal's identity is consistent. For (nk)+(nk1)=(n+1k), the larger lower index is k in the result.

Step 3: Final Application of Pascal's Identity and Determine m○ Expand

Apply Pascal's identity one last time to the simplified expression:

(32r)+(32r1)=(33r)

Given that the original expression equals (mr), we have:

(33r)=(mr)

Comparing both sides, we find that m=33.

💡 Teacher's Secret Hint

The range 3r30 ensures all binomial coefficients are well-defined and valid for the application of these identities.

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