Find sum_{r=1}^{5} C(5, r) | vedclass

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Question

(B) The expression is $\sum_{r=1}^{5} C(5, r) = C(5, 1) + C(5, 2) + C(5, 3) + C(5, 4) + C(5, 5)$.We know that the sum of binomial coefficients is $\su

Vedclass · Accepted Answer

$31$

Vedclass · Answer

(B) The expression is $\sum_{r=1}^{5} C(5, r) = C(5, 1) + C(5, 2) + C(5, 3) + C(5, 4) + C(5, 5)$.We know that the sum of binomial coefficients is $\sum_{r=0}^{n} C(n, r) = 2^n$.Here,$n = 5$,so $\sum_{r=0}^{5} C(5, r) = 2^5 = 32$.The given sum is $\sum_{r=1}^{5} C(5, r) = \left( \sum_{r=0}^{5} C(5, r) \right) - C(5, 0)$.Since $C(5, 0) = 1$,we have $32 - 1 = 31$.Thus,the correct option is $B$.

Find $\sum_{r=1}^{5} C(5, r)$

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