If 1^2 * (15C1) + 2^2 * (15C2) + 3^2 * (15C3) + ... + 15^2 * (15C15) = 2^m * 3^n * 5^k, where m, n, k ∈ N, then m+n+k is equal to :

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q7 – JEE-MAIN 2025

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Home Maths Binomial TheoremQuestion #48

1^{2} \cdot (^{15} C_{1}) + 2^{2} \cdot (^{15} C_{2}) + 3^{2} \cdot (^{15} C_{3}) + . . . + 15^{2} \cdot (^{15} C_{15}) = 2^{m} \cdot 3^{n} \cdot 5^{k}

, where

m, n, k \in N

, then

m + n + k

is equal to :

Clue 1 of 3: Binomial Series Summation

Recognize the given series as a sum involving binomial coefficients and squared indices.

🧠 Full Solution Path

Step 1: Identify the Summation Series✦ Active

The given series is $S = 1^{2} \cdot (^{15} C_{1}) + 2^{2} \cdot (^{15} C_{2}) + 3^{2} \cdot (^{15} C_{3}) + . . . + 15^{2} \cdot (^{15} C_{15})$ . This can be written in summation notation as $S = \sum_{r = 1}^{15} r^{2} \cdot^{15} C_{r}$ . Since $0^{2} \cdot^{15} C_{0} = 0$ , the sum is equivalent to $\sum_{r = 0}^{15} r^{2} \cdot^{15} C_{r}$ .

Step 2: Apply Binomial Coefficient Identity○ Expand

We use the standard identity for the sum of $r^{2} \cdot^{n} C_{r}$ : $\sum_{r = 0}^{n} r^{2} \cdot^{n} C_{r} = n (n + 1) 2^{n - 2}$ . For this problem, $n = 15$ . Substituting $n = 15$ into the identity:

S = 15 (15 + 1) 2^{15 - 2}

Calculate the value:

S = 15 \cdot 16 \cdot 2^{13}

Since $16 = 2^{4}$ , we have:

S = 15 \cdot 2^{4} \cdot 2^{13} = 15 \cdot 2^{17}

💡 Teacher's Secret Hint

Remember that $r^{2} \cdot^{n} C_{r} = r \cdot n \cdot^{n - 1} C_{r - 1} = ((r - 1) + 1) \cdot n \cdot^{n - 1} C_{r - 1}$ can be used to derive this identity if you don't recall it directly.

Step 3: Determine m, n, k and Calculate m+n+k○ Expand

The calculated sum is $S = 15 \cdot 2^{17}$ . We need to express this in the form $2^{m} \cdot 3^{n} \cdot 5^{k}$ . Prime factorize 15 as $3 \cdot 5$ :

S = (3^{1} \cdot 5^{1}) \cdot 2^{17} = 2^{17} \cdot 3^{1} \cdot 5^{1}

Comparing this with $2^{m} \cdot 3^{n} \cdot 5^{k}$ , we get $m = 17$ , $n = 1$ , and $k = 1$ . All are natural numbers as required. Finally, calculate $m + n + k$ :

m + n + k = 17 + 1 + 1 = 19

💡 Teacher's Secret Hint

Ensure all exponents are positive integers, as $m, n, k \in N$ (natural numbers).

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Maths Question 7 – JEE-MAIN 2025

Clue 1 of 3: Binomial Series Summation

🧠 Full Solution Path