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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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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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

Logic and Functions

Decision Maths

Geometry

Probability and Statistics

Calculus

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

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