Chapter 1: Q. 1.9 (page 17)

Use Theoretical Exercise 8 to prove that
$(\begin{matrix} 2 n \\ n \end{matrix}) = \sum_{k = 0}^{n} {(\begin{matrix} n \\ k \end{matrix})}^{2}$

Short Answer

Expert verified

It is proved that $(\begin{matrix} 2 n \\ n \end{matrix}) = \sum_{k = 0}^{n} {(\begin{matrix} n \\ k \end{matrix})}^{2}$

Step by step solution

Step 1. State the Theoretical Exercise 8.

According to Theoretical Exercise 8,

$(\begin{matrix} n + m \\ r \end{matrix}) = (\begin{matrix} n \\ 0 \end{matrix}) (\begin{matrix} m \\ r \end{matrix}) + (\begin{matrix} n \\ 1 \end{matrix}) (\begin{matrix} m \\ r - 1 \end{matrix}) + . . . . + (\begin{matrix} n \\ r \end{matrix}) (\begin{matrix} m \\ 0 \end{matrix})$ ...................... (1)

Step 2. Prove the given equation.

By substituting $m = n$ and $r = n$ in equation (1), we get

role="math" localid="1647934551213" $(\begin{matrix} n + n \\ n \end{matrix}) = (\begin{matrix} n \\ 0 \end{matrix}) (\begin{matrix} n \\ n \end{matrix}) + (\begin{matrix} n \\ 1 \end{matrix}) (\begin{matrix} n \\ n - 1 \end{matrix}) + . . . . + (\begin{matrix} n \\ n \end{matrix}) (\begin{matrix} n \\ 0 \end{matrix})$

role="math" localid="1647934685591" $(\begin{matrix} 2 n \\ n \end{matrix}) = (\begin{matrix} n \\ 0 \end{matrix}) (\begin{matrix} n \\ n \end{matrix}) + (\begin{matrix} n \\ 1 \end{matrix}) (\begin{matrix} n \\ n - 1 \end{matrix}) + . . . . + (\begin{matrix} n \\ n \end{matrix}) (\begin{matrix} n \\ 0 \end{matrix})$ .......................... (2)

We know that,

${}^{n}C_{r} = {}^{n}C_{n - r}$ ............................. (3)

Substituting (3) in (2), we get

$(\begin{matrix} 2 n \\ n \end{matrix}) = {(\begin{matrix} n \\ 0 \end{matrix})}^{2} + {(\begin{matrix} n \\ 1 \end{matrix})}^{2} + . . . . + {(\begin{matrix} n \\ n \end{matrix})}^{2}$

Therefore, it is proved that $(\begin{matrix} 2 n \\ n \end{matrix}) = \sum_{k = 0}^{n} {(\begin{matrix} n \\ k \end{matrix})}^{2}$ .

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Use Theoretical Exercise 8 to prove that
$(\begin{matrix} 2 n \\ n \end{matrix}) = \sum_{k = 0}^{n} {(\begin{matrix} n \\ k \end{matrix})}^{2}$

Short Answer

Step by step solution

Step 1. State the Theoretical Exercise 8.

Step 2. Prove the given equation.

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Use Theoretical Exercise 8 to prove that2nn=∑k=0nnk2

Short Answer

Step by step solution

Step 1. State the Theoretical Exercise 8.

Step 2. Prove the given equation.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Applied Mathematics

Logic and Functions

Pure Maths

Discrete Mathematics

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Use Theoretical Exercise 8 to prove that
$(\begin{matrix} 2 n \\ n \end{matrix}) = \sum_{k = 0}^{n} {(\begin{matrix} n \\ k \end{matrix})}^{2}$