If {}^n C_0, {}^n C_1, {}^n C_2, ldots, {}^n | vedclass

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(B) Given $n=10$,we need to evaluate $S = \sum_{r=1}^{10} {}^n C_r r(r-4)$.Expanding the term: $r(r-4) = r(r-1) - 3r$.So,$S = \sum_{r=1}^{10} {}^n C_r

Vedclass · Accepted Answer

$7680$

Vedclass · Answer

(B) Given $n=10$,we need to evaluate $S = \sum_{r=1}^{10} {}^n C_r r(r-4)$.Expanding the term: $r(r-4) = r(r-1) - 3r$.So,$S = \sum_{r=1}^{10} {}^n C_r r(r-1) - 3 \sum_{r=1}^{10} r {}^n C_r$.Using the identities $r(r-1) {}^n C_r = n(n-1) {}^{n-2} C_{r-2}$ and $r {}^n C_r = n {}^{n-1} C_{r-1}$:$S = n(n-1) \sum_{r=2}^{10} {}^{n-2} C_{r-2} - 3n \sum_{r=1}^{10} {}^{n-1} C_{r-1}$.Since $\sum_{k=0}^{m} {}^m C_k = 2^m$,we have:$S = n(n-1) 2^{n-2} - 3n 2^{n-1}$.Substituting $n=10$:$S = 10(9) 2^8 - 3(10) 2^9 = 90 \cdot 2^8 - 30 \cdot 2 \cdot 2^8$.$S = 90 \cdot 2^8 - 60 \cdot 2^8 = 30 \cdot 2^8$.$S = 30 	imes 256 = 7680$.

If ${}^n C_0, {}^n C_1, {}^n C_2, \ldots, {}^n C_n$ are the binomial coefficients in the expansion of $(1+x)^n$,then for $n=10$,the value of $\sum_{r=1}^{10} {}^n C_r \cdot r(r-4)$ is:

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