If sumlimits_{i = 1}^{20} {left( {frac{{{}^{2 | vedclass

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(D) We know the property of binomial coefficients: ${}^{n}{C_r} + {}^{n}{C_{r-1}} = {}^{n+1}{C_r}$.Applying this to the denominator: ${}^{20}{C_i} + {

Vedclass · Accepted Answer

$100$

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(D) We know the property of binomial coefficients: ${}^{n}{C_r} + {}^{n}{C_{r-1}} = {}^{n+1}{C_r}$.Applying this to the denominator: ${}^{20}{C_i} + {}^{20}{C_{i-1}} = {}^{21}{C_i}$.Thus,the term inside the summation is $\frac{{}^{20}{C_{i-1}}}{{}^{21}{C_i}}$.Using the formula ${}^{n}{C_r} = \frac{n}{r} \cdot {}^{n-1}{C_{r-1}}$,we have ${}^{21}{C_i} = \frac{21}{i} \cdot {}^{20}{C_{i-1}}$.Substituting this into the expression: $\frac{{}^{20}{C_{i-1}}}{\frac{21}{i} \cdot {}^{20}{C_{i-1}}} = \frac{i}{21}$.The summation becomes $\sum\limits_{i = 1}^{20} {\left( \frac{i}{21} ight)}^3 = \frac{1}{21^3} \sum\limits_{i = 1}^{20} i^3$.Using the sum of cubes formula $\sum\limits_{i=1}^{n} i^3 = \left( \frac{n(n+1)}{2} ight)^2$,for $n=20$:$\sum\limits_{i=1}^{20} i^3 = \left( \frac{20 	imes 21}{2} ight)^2 = (10 	imes 21)^2 = 100 	imes 21^2$.Substituting back: $S = \frac{100 	imes 21^2}{21^3} = \frac{100}{21}$.Given $S = \frac{k}{21}$,we find $k = 100$.

If $\sum\limits_{i = 1}^{20} {\left( {\frac{{{}^{20}{C_{i - 1}}}}{{{}^{20}{C_i} + {}^{20}{C_{i - 1}}}}} \right)} ^3 = \frac{k}{21}$,then $k$ equals

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