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7. (+1) (n+3 (n+3)(n+2) 2 for n ≥ -1

7. (+1) (n+3 (n+3)(n+2) 2 for n ≥ -1

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 33EQ
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Can you please help with #7 and #9? TIA

Justify the equations in 6-9 either by deriving them from for- mulas in Example 9.7.1 or by direct computation from Theo- rem 9.5.1. Assume m, n, k, and r are integers. 6. (m+k₁) - = m + k, for m+k≥ 1 1 n+3 7. (+³) = (n+3)(n+2) 2 , for n ≥ -1 n+1 8. (7) = 1, for k-r≥0 k r 2n 9. (²2) for n ≥ 0 n
Transcribed Image Text:Justify the equations in 6-9 either by deriving them from for- mulas in Example 9.7.1 or by direct computation from Theo- rem 9.5.1. Assume m, n, k, and r are integers. 6. (m+k₁) - = m + k, for m+k≥ 1 1 n+3 7. (+³) = (n+3)(n+2) 2 , for n ≥ -1 n+1 8. (7) = 1, for k-r≥0 k r 2n 9. (²2) for n ≥ 0 n
Theorem 9.7.1 Pascal's Formula Let n and r be positive integers and suppose r ≤n. Then n+ (" + ¹) = ( ² ) + ( ) · r Proof (algebraic version): Let n and r be positive integers with r ≤n. By Theorem 9.5.1, n! ( ²₁ ) + ( )= + (r − 1)!(n − (r − 1))! - r!(n-r)! n! n! + (r − 1)!(n − r + 1)!r!(n-r)!* To add these fractions, a common denominator is needed, so multiply the numera- tor and denominator of the left-hand fraction by r and multiply the numerator and denominator of the right-hand fraction by (n-r+1). Then n! r n! (²₁) + ( ) (r − 1)!(n −r + 1)! r (n-r+1) r!(n-r)! (nr+1) n-n! -n!-r+n! n!.r (n-r+1)!r(r 1)! (nr+1)(n-r)!r! _n!+r+n!•n__n!r+n! = n!(n + 1) (n+1-r)!r! (n-r+1)!r! (n + 1)! = ((n + 1) -r)!r! = = ("+¹).
Transcribed Image Text:Theorem 9.7.1 Pascal's Formula Let n and r be positive integers and suppose r ≤n. Then n+ (" + ¹) = ( ² ) + ( ) · r Proof (algebraic version): Let n and r be positive integers with r ≤n. By Theorem 9.5.1, n! ( ²₁ ) + ( )= + (r − 1)!(n − (r − 1))! - r!(n-r)! n! n! + (r − 1)!(n − r + 1)!r!(n-r)!* To add these fractions, a common denominator is needed, so multiply the numera- tor and denominator of the left-hand fraction by r and multiply the numerator and denominator of the right-hand fraction by (n-r+1). Then n! r n! (²₁) + ( ) (r − 1)!(n −r + 1)! r (n-r+1) r!(n-r)! (nr+1) n-n! -n!-r+n! n!.r (n-r+1)!r(r 1)! (nr+1)(n-r)!r! _n!+r+n!•n__n!r+n! = n!(n + 1) (n+1-r)!r! (n-r+1)!r! (n + 1)! = ((n + 1) -r)!r! = = ("+¹).
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Thank you. Can you please explain why you multiplied by 2! in the denominator in step 2? How does that 2! get there?

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