If 3 times { }^5 C_0 + 8 times { }^5 C_1 + 13 | vedclass

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(C) The given expression is $\sum_{r=0}^{5} (5r + 3) 	imes { }^5 C_r$.We know that $\sum_{r=0}^{n} { }^n C_r = 2^n$ and $\sum_{r=0}^{n} r 	imes { }^

Vedclass · Accepted Answer

$31$

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(C) The given expression is $\sum_{r=0}^{5} (5r + 3) 	imes { }^5 C_r$.We know that $\sum_{r=0}^{n} { }^n C_r = 2^n$ and $\sum_{r=0}^{n} r 	imes { }^n C_r = n 	imes 2^{n-1}$.Here,$n = 5$.So,$\sum_{r=0}^{5} (5r + 3) 	imes { }^5 C_r = 5 \sum_{r=0}^{5} r 	imes { }^5 C_r + 3 \sum_{r=0}^{5} { }^5 C_r$.Substituting the values:$= 5 	imes (5 	imes 2^{5-1}) + 3 	imes 2^5$$= 5 	imes (5 	imes 2^4) + 3 	imes (2 	imes 2^4)$$= 25 	imes 2^4 + 6 	imes 2^4$$= (25 + 6) 	imes 2^4 = 31 	imes 2^4$.Given $k 	imes 2^4 = 31 	imes 2^4$,therefore $k = 31$.

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