Q.48 If n is a positive integer, show... [FREE SOLUTION]

91影视

Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 12: Q.48 (page 837)

If n is a positive integer, show that
$(\begin{matrix} n \\ 0 \end{matrix}) - (\begin{matrix} n \\ 1 \end{matrix}) + (\begin{matrix} n \\ 2 \end{matrix}) - (\begin{matrix} n \\ 3 \end{matrix}) + . . . . . . + {(- 1)}^{n} (\begin{matrix} n \\ n \end{matrix}) = 0$

Short Answer

Expert verified

The Binomial theorem states that

Step by step solution

Step 1. Given information

In this question $(\begin{matrix} n \\ 0 \end{matrix}) - (\begin{matrix} n \\ 1 \end{matrix}) + (\begin{matrix} n \\ 2 \end{matrix}) + . . . . . . . . + {(- 1)}^{n} (\begin{matrix} n \\ n \end{matrix}) i s g i v e n$

We have to prove that $(\begin{matrix} n \\ 0 \end{matrix}) - (\begin{matrix} n \\ 1 \end{matrix}) + (\begin{matrix} n \\ 2 \end{matrix}) + . . . . . . . . + {(- 1)}^{n} (\begin{matrix} n \\ n \end{matrix}) = 0$

. Description of proving the given question

By using the binomial theorem we can prove this given statement

According to the Binomial theorem

$L e t x a n d a b e r e a l n u m b e r s . F o r a n y p o s i t i v e i n t e g e r n, w e h a v e {(x + a)}^{n} = (\begin{matrix} n \\ 0 \end{matrix}) x^{n} + (\begin{matrix} n \\ 1 \end{matrix}) a x^{n - 1} + (\begin{matrix} n \\ 2 \end{matrix}) a^{2} x^{n - 2} + . . . . . . . . . . . . + (\begin{matrix} n \\ n \end{matrix}) a^{n} p u t x = 1 a n d a = - 1 t h e n b y a p p l y i n g t h e o r e m, w e c a n e x p a n d {(1 - 1)}^{n} {(1 - 1)}^{n} = (\begin{matrix} n \\ 0 \end{matrix}) 1^{n} + (\begin{matrix} n \\ 1 \end{matrix}) (- 1) 1^{n - 1} + (\begin{matrix} n \\ 2 \end{matrix}) {(- 1)}^{2} 1^{n - 2} + . . . . . . . . . . . . + (\begin{matrix} n \\ n \end{matrix}) {(- 1)}^{n} 1^{n - n} = (\begin{matrix} n \\ 0 \end{matrix}) - (\begin{matrix} n \\ 1 \end{matrix}) + (\begin{matrix} n \\ 2 \end{matrix}) + . . . . . . . . . . . . + {(- 1)}^{n} (\begin{matrix} n \\ n \end{matrix}) b u t {(1 - 1)}^{n} = {(0)}^{n} \sin c e (1 - 1) = 0 = 0 h e n c e (\begin{matrix} n \\ 0 \end{matrix}) - (\begin{matrix} n \\ 1 \end{matrix}) + (\begin{matrix} n \\ 2 \end{matrix}) + . . . . . . . . . . . . + {(- 1)}^{n} (\begin{matrix} n \\ n \end{matrix}) = 0$

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91影视

If n is a positive integer, show that
$(\begin{matrix} n \\ 0 \end{matrix}) - (\begin{matrix} n \\ 1 \end{matrix}) + (\begin{matrix} n \\ 2 \end{matrix}) - (\begin{matrix} n \\ 3 \end{matrix}) + . . . . . . + {(- 1)}^{n} (\begin{matrix} n \\ n \end{matrix}) = 0$

Short Answer

Step by step solution

Step 1. Given information

. Description of proving the given question

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91影视

If n is a positive integer, show thatn0-n1+n2-n3+......+(-1)nnn=0

Short Answer

Step by step solution

Step 1. Given information

. Description of proving the given question

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Logic and Functions

Decision Maths

Calculus

Theoretical and Mathematical Physics

Applied Mathematics

Study anywhere. Anytime. Across all devices.

If n is a positive integer, show that
$(\begin{matrix} n \\ 0 \end{matrix}) - (\begin{matrix} n \\ 1 \end{matrix}) + (\begin{matrix} n \\ 2 \end{matrix}) - (\begin{matrix} n \\ 3 \end{matrix}) + . . . . . . + {(- 1)}^{n} (\begin{matrix} n \\ n \end{matrix}) = 0$