### Related Formula The general property linking indices to binomial coefficients is: r cdot binomnr = n cdot binomn-1r-1$r \cdot \binom{n}{r} = n \cdot \binom{n-1}{r-1}$ ### Core Logic The given series can be structured using summation notation: S = sum_r=1^15 r^2 cdot binom15r$S = \sum_{r=1}^{15} r^2 \cdot \binom{15}{r}$ Apply the identity r cdot binom15r = 15 cdot binom14r-1$r \cdot \binom{15}{r} = 15 \cdot \binom{14}{r-1}$ to reduce one factor of r$r$: S = sum_r=1^15 r cdot left[ 15 cdot binom14r-1 right] = 15 sum_r=1^15 r cdot binom14r-1$S = \sum_{r=1}^{15} r \cdot \left[ 15 \cdot \binom{14}{r-1} \right] = 15 \sum_{r=1}^{15} r \cdot \binom{14}{r-1}$ ### Step 1: Splitting the linear term Rewrite the index variable r$r$ as (r - 1) + 1$(r - 1) + 1$ to align with the binomial lower index: S = 15 sum_r=1^15 big((r - 1) + 1big) cdot binom14r-1$S = 15 \sum_{r=1}^{15} \big((r - 1) + 1\big) \cdot \binom{14}{r-1}$ S = 15 sum_r=1^15 (r - 1) cdot binom14r-1 + 15 sum_r=1^15 binom14r-1$S = 15 \sum_{r=1}^{15} (r - 1) \cdot \binom{14}{r-1} + 15 \sum_{r=1}^{15} \binom{14}{r-1}$ Applying the property again to the first summation term: (r-1)binom14r-1 = 14binom13r-2$(r-1)\binom{14}{r-1} = 14\binom{13}{r-2}$: S = 15 cdot 14 sum_r=2^15 binom13r-2 + 15 sum_r=1^15 binom14r-1$S = 15 \cdot 14 \sum_{r=2}^{15} \binom{13}{r-2} + 15 \sum_{r=1}^{15} \binom{14}{r-1}$ ### Step 2: Evaluating the Sums and Prime Factorization Using the standard total sum of binomial coefficients sum_k=0^n binomnk = 2^n$\sum_{k=0}^{n} \binom{n}{k} = 2^n$: S = 15 cdot 14 cdot 2^13 + 15 cdot 2^14$S = 15 \cdot 14 \cdot 2^{13} + 15 \cdot 2^{14}$ Factor out 15 cdot 2^13$15 \cdot 2^{13}$ from the expression: S = 15 cdot 2^13 (14 + 2) = 15 cdot 2^13 (16) = 15 cdot 2^13 cdot 2^4$S = 15 \cdot 2^{13} (14 + 2) = 15 \cdot 2^{13} (16) = 15 \cdot 2^{13} \cdot 2^4$ S = 15 cdot 2^17 = (3^1 cdot 5^1) cdot 2^17$S = 15 \cdot 2^{17} = (3^1 \cdot 5^1) \cdot 2^{17}$ Matching this with the given format 2^m cdot 3^n cdot 5^k$2^m \cdot 3^n \cdot 5^k$, we identify: m = 17$m = 17$, n = 1$n = 1$, and k = 1$k = 1$. ### Step 3: Calculating the sum of exponents Evaluating the targeted summation: m + n + k = 17 + 1 + 1 = 19$m + n + k = 17 + 1 + 1 = 19$ ### Pattern Recognition For a series of the type sum r^2 binomnr$\sum r^2 \binom{n}{r}$, remember the standard identity shortcut: n(n-1)2^n-2 + n2^n-1$n(n-1)2^{n-2} + n2^{n-1}$. Plugging in n=15$n=15$ instantly outputs 15(14)2^13 + 15(2^14)$15(14)2^{13} + 15(2^{14})$, bypasses matching terms manually. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem