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Question
For all positive integers n, show that 2nCn + 2nCn − 1 = `1/2` 2n + 2Cn+1
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Solution
\[LHS = {}^{2n} C_n + {}^{2n} C_{n - 1} \]
\[ = \frac{\left( 2n \right)!}{n! n!} + \frac{\left( 2n \right)!}{\left( n - 1 \right)! \left( 2n - n + 1 \right)!}\]
\[ = \frac{\left( 2n \right)!}{n! n!} + \frac{\left( 2n \right)!}{\left( n - 1 \right)! \left( n + 1 \right)!}\]
\[ = \frac{\left( 2n \right)!}{n \left( n - 1 \right)! n!} + \frac{\left( 2n \right)!}{\left( n - 1 \right)! \left( n + 1 \right)n!}\]
\[ = \frac{\left( 2n \right)!}{n! \left( n - 1 \right)!} \left[ \frac{1}{n} + \frac{1}{n + 1} \right]\]
\[ = \frac{\left( 2n \right)!}{n! \left( n - 1 \right)!} \left[ \frac{2n + 1}{n \left( n + 1 \right)} \right]\]
\[ = \frac{\left( 2n + 1 \right)!}{n! \left( n + 1 \right)!}\]
\[RHS = \frac{1}{2} {}^{2n + 2} C_{n + 1} \]
\[ = \frac{1}{2} \left[ \frac{\left( 2n + 2 \right)!}{\left( n + 1 \right)! \left( 2n + 2 - n - 1 \right)!} \right]\]
\[ = \frac{1}{2} \left[ \frac{\left( 2n + 2 \right)!}{\left( n + 1 \right)! \left( n + 1 \right)!} \right]\]
\[ = \frac{1}{2} \left[ \frac{\left( 2n + 2 \right) \left( 2n + 1 \right)!}{\left( n + 1 \right) n! \left( n + 1 \right)!} \right]\]
\[ = \frac{1}{2} \left[ \frac{2\left( n + 1 \right) \left( 2n + 1 \right)!}{\left( n + 1 \right) n! \left( n + 1 \right)!} \right]\]
\[ = \frac{\left( 2n + 1 \right)!}{n! \left( n + 1 \right)!}\]
∴ LHS = RHS
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