# For all positive integers n, show that 2nCn + 2nCn − 1 = 12 2n + 2Cn+1 - Mathematics

Sum

For all positive integers n, show that 2nCn + 2nCn − 1 = 1/2 2n + 2Cn+1

#### 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

Concept: Factorial N (N!) Permutations and Combinations
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 17 Combinations
Exercise 17.1 | Q 17 | Page 8