sumlimits_{k = 0}^{10} {^{20}{C_k} = } | vedclass

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(A) Let $S = \sum\limits_{k = 0}^{10} {^{20}C_k} = {^{20}C_0} + {^{20}C_1} + \dots + {^{20}C_{10}}$.We know that $\sum\limits_{k = 0}^{20} {^{20}C_k}

Vedclass · Accepted Answer

$2^{19} + \frac{1}{2} {^{20}C_{10}}$

Vedclass · Answer

(A) Let $S = \sum\limits_{k = 0}^{10} {^{20}C_k} = {^{20}C_0} + {^{20}C_1} + \dots + {^{20}C_{10}}$.We know that $\sum\limits_{k = 0}^{20} {^{20}C_k} = 2^{20}$.Since ${^{n}C_r} = {^{n}C_{n-r}}$,we have ${^{20}C_0} = {^{20}C_{20}}$,${^{20}C_1} = {^{20}C_{19}}$,...,${^{20}C_9} = {^{20}C_{11}}$.Thus,$2^{20} = ({^{20}C_0} + {^{20}C_1} + \dots + {^{20}C_9}) + {^{20}C_{10}} + ({^{20}C_{11}} + \dots + {^{20}C_{20}})$.$2^{20} = 2({^{20}C_0} + {^{20}C_1} + \dots + {^{20}C_9}) + {^{20}C_{10}}$.Let $S = {^{20}C_0} + {^{20}C_1} + \dots + {^{20}C_9} + {^{20}C_{10}}$.Then $2S = 2({^{20}C_0} + \dots + {^{20}C_9}) + 2{^{20}C_{10}}$.Substituting the sum,$2S = (2^{20} - {^{20}C_{10}}) + 2{^{20}C_{10}} = 2^{20} + {^{20}C_{10}}$.Therefore,$S = 2^{19} + \frac{1}{2} {^{20}C_{10}}$.

$\sum\limits_{k = 0}^{10} {^{20}{C_k} = }$

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