JEE Mains · Maths · STD 11 - 7. binomial theoram
\(\left( {\left( {\begin{array}{*{20}{c}}
{21}\\
1
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
1
\end{array}} \right)} \right) + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
2
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
2
\end{array}} \right)} \right)\)\( + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
3
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
3
\end{array}} \right)} \right) + \;.\;.\;.\)\( + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
{10}
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
{10}
\end{array}} \right)} \right) = \)
- A \({2^{20}} - {2^{10}}\)
- B \({2^{21}} - {2^{11}}\)
- C \({2^{21}} - {2^{10}}\)
- D \({2^{20}} - {2^9}\)
Answer & Solution
Correct Answer
(A) \({2^{20}} - {2^{10}}\)
Step-by-step Solution
Detailed explanation
We have \(\left(^{21} \mathrm{C}_{1}+^{21} \mathrm{C}_{2} \ldots \ldots+^{21} \mathrm{C}_{10}\right)\) \(-\left(^{10} \mathrm{C}_{1}+^{10} \mathrm{C}_{2} \ldots . .^{10} \mathrm{C}_{10}\right)\)…
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