Evaluate the sum: left( binom{21}{1} - binom{ | vedclass

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Question

(A) The given expression is $S = \sum_{r=1}^{10} \binom{21}{r} - \sum_{r=1}^{10} \binom{10}{r}$.We know that $\sum_{r=0}^{21} \binom{21}{r} = 2^{21}$.

Vedclass · Accepted Answer

$2^{20} - 2^{10}$

Vedclass · Answer

(A) The given expression is $S = \sum_{r=1}^{10} \binom{21}{r} - \sum_{r=1}^{10} \binom{10}{r}$.We know that $\sum_{r=0}^{21} \binom{21}{r} = 2^{21}$.Since $\binom{21}{r} = \binom{21}{21-r}$,we have $\sum_{r=0}^{10} \binom{21}{r} = \sum_{r=11}^{21} \binom{21}{r}$.Thus,$2 \sum_{r=1}^{10} \binom{21}{r} + \binom{21}{0} + \binom{21}{11} = 2^{21}$ is not quite right; rather,$2 \sum_{r=1}^{10} \binom{21}{r} + \binom{21}{0} + \binom{21}{21} = 2^{21}$ is incorrect. Correctly,$\sum_{r=0}^{21} \binom{21}{r} = 2^{21} \implies 2 \sum_{r=0}^{10} \binom{21}{r} = 2^{21} \implies \sum_{r=0}^{10} \binom{21}{r} = 2^{20}$.Therefore,$\sum_{r=1}^{10} \binom{21}{r} = 2^{20} - \binom{21}{0} = 2^{20} - 1$.Next,$\sum_{r=1}^{10} \binom{10}{r} = (1+1)^{10} - \binom{10}{0} = 2^{10} - 1$.Substituting these into the original expression:$S = (2^{20} - 1) - (2^{10} - 1) = 2^{20} - 2^{10}$.

Evaluate the sum: $\left( \binom{21}{1} - \binom{10}{1} \right) + \left( \binom{21}{2} - \binom{10}{2} \right) + \left( \binom{21}{3} - \binom{10}{3} \right) + \dots + \left( \binom{21}{10} - \binom{10}{10} \right) = $

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