how would one approche these problems here stuck. (b)x*yn-k is the same as the coefficient in front of x"-kyk.Consider the expression (x + y)", show that the coefficient in front of (b)Consider the sequence (xn)n, n E NU{0} defined recursively asXn = 7xn-1 – 10xn-2;xo = 2, x13.Show that the nth term is given by the closed formula1(7 · 2" – 3").Xn

Answered: (b) x*yn-k is the same as the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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how would one approche these problems here stuck.

(b)
x*yn-k is the same as the coefficient in front of x"-kyk.
Consider the expression (x + y)", show that the coefficient in front of

(b)
Consider the sequence (xn)n, n E NU{0} defined recursively as
Xn = 7xn-1 – 10xn-2;
xo = 2, x1
3.
Show that the nth term is given by the closed formula
1
(7 · 2" – 3").
Xn

Expert Solution

Step 1

As per Bartleby guidelines, for more than one questions are asked, first is to be answered. Please upload the other separately.

Consider the binomial expansion of ${(x + y)}^{n}$ is,

${(x + y)}^{n} = \sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) x^{n - k} y^{k}$

To prove that the coefficient of $x^{k} y^{n - k}$ and the coefficient of $x^{n - k} y^{k}$ are same.

From the above binomial expansion,

The coefficient of $x^{n - k} y^{k}$ is $(\begin{matrix} n \\ k \end{matrix})$ and the coefficient of $x^{k} y^{n - k}$ is $(\begin{matrix} n \\ n - k \end{matrix})$ .

Therefore, it is enough to prove that $(\begin{matrix} n \\ k \end{matrix}) = (\begin{matrix} n \\ n - k \end{matrix})$

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