If 1^2 \cdot( ^15 C_1)+2^2 \cdot( ^15 C_2)+3^2 \cdot( ^15 C_3)+\ldots+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 :

Binomial Theorem PYQ — JEE Main 2025 · 4 Apr · Shift 2 · Q38 with Solution

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Correct answer: B

\sum_{r=1}^{15} r^2 {15\choose r}.$$ 2. Use the identity $$r^2=r(r-1)+r.$$ So, $$\sum_{r=0}^{15} r^2{15\choose r}=\sum_{r=0}^{15} r(r-1){15\choose r}+\sum_{r=0}^{15} r{15\choose r}.$$ 3. Now use standard binomial sum identities: $$\sum_{r=0}^{n} r{n\choose r}=n2^{n-1},$$ $$\sum_{r=0}^{n} r(r-1){n\choose r}=n(n-1)2^{n-2}.$$ For $n=15$, $$\sum_{r=0}^{15} r{15\choose r}=15\cdot 2^{14},$$ $$\sum_{r=0}^{15} r(r-1){15\choose r}=15\cdot 14\cdot 2^{13}.$$ 4. Therefore, $$S=15\cdot 14\cdot 2^{13}+15\cdot 2^{14}.$$ Factor this: $$S=15\cdot 2^{13}(14+2)=15\cdot 2^{13}\cdot 16=15\cdot 2^{17}.$$ Now, $$15=3\cdot 5,$$ so $$S=2^{17}\cdot 3^1\cdot 5^1.$$ 5. Hence, $$m=17,\quad n=1,\quad k=1.$$ Therefore, $$m+n+k=17+1+1=19.$$ 6. Checking options: the correct option is **B: 19**.

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Correct answer: B

\sum_{r=1}^{15} r^2 {15\choose r}.$$ 2. Use the identity $$r^2=r(r-1)+r.$$ So, $$\sum_{r=0}^{15} r^2{15\choose r}=\sum_{r=0}^{15} r(r-1){15\choose r}+\sum_{r=0}^{15} r{15\choose r}.$$ 3. Now use standard binomial sum identities: $$\sum_{r=0}^{n} r{n\choose r}=n2^{n-1},$$ $$\sum_{r=0}^{n} r(r-1){n\choose r}=n(n-1)2^{n-2}.$$ For $n=15$, $$\sum_{r=0}^{15} r{15\choose r}=15\cdot 2^{14},$$ $$\sum_{r=0}^{15} r(r-1){15\choose r}=15\cdot 14\cdot 2^{13}.$$ 4. Therefore, $$S=15\cdot 14\cdot 2^{13}+15\cdot 2^{14}.$$ Factor this: $$S=15\cdot 2^{13}(14+2)=15\cdot 2^{13}\cdot 16=15\cdot 2^{17}.$$ Now, $$15=3\cdot 5,$$ so $$S=2^{17}\cdot 3^1\cdot 5^1.$$ 5. Hence, $$m=17,\quad n=1,\quad k=1.$$ Therefore, $$m+n+k=17+1+1=19.$$ 6. Checking options: the correct option is **B: 19**.