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$\left( {\left( {\begin{array}{*{20}{c}}
{21}\\
1
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
1
\end{array}} \right)} \right) + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
2
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
2
\end{array}} \right)} \right)$$ + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
3
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
3
\end{array}} \right)} \right) + \;.\;.\;.$$ + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
{10}
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
{10}
\end{array}} \right)} \right)$ का मान है:
${2^{20}} - {2^{10}}$
${2^{21}} - {2^{11}}$
${2^{21}} - {2^{10}}$
${2^{20}} - {2^9}$
Solution
We have $\left(^{21} \mathrm{C}_{1}+^{21} \mathrm{C}_{2} \ldots \ldots+^{21} \mathrm{C}_{10}\right)$
$-\left(^{10} \mathrm{C}_{1}+^{10} \mathrm{C}_{2} \ldots . .^{10} \mathrm{C}_{10}\right)$
$=\frac{1}{2}\left[\left(^{21} \mathrm{C}_{1}+\ldots+^{21} \mathrm{C}_{10}\right)+\left(^{21} \mathrm{C}_{11}+\ldots^{21} \mathrm{C}_{20}\right)\right]-\left(2^{10}-1\right)$
$\left(\because 10 \mathrm{C}_{1}+^{10} \mathrm{C}_{2}+\ldots .+^{10} \mathrm{C}_{10}=2^{10}-1\right)$
$=\frac{1}{2}\left[2^{21}-2\right]-\left(2^{10}-1\right)$
$=\left(2^{20}-1\right)-\left(2^{10}-1\right)=2^{20}-2^{10}$
