Compute the values: D2-2D1, D3-3D2, D4-4D3, D5-5D4, …, D10-10D9, where the D’s are the derangements values described on P.335. What is the pattern? Using Eq. 8.1.2 together with the values of D, and D2, we can evaluate D, forany value of n:// in principleD3D4D5D6D7DsD9D10(3 – 1){D2+D1} = 2{1+0}(4 – 1){D3 +D2} = 3{2+1}(5 – 1){D4+D3} = 4{9+2}(6 – 1){D5 +D4} = 5{44+9}(7 – 1){D6+D5} = 6{265 +44}(8 – 1){D7+D6} = 7{1854 +265}(9 – 1){Ds +D7} = 8{14833 + 1854}(10 – 1){D9+Ds} = 9{133496 +14833}2-%3!442651 85414 833-133 4961 334 961// It's strange that 1 334 961 = 10× (133 496) + 1. Or is it?// Is there a (convenient and compact) formula for D, that we can use to calculate// its values?|| || |||| | ||||||

Here is the solution:- Now, according to the given information D2-2D1, D3-3D2, D4-4D3, D5-5D4, …, D10-10D9, D1=0,D2=1,D3=2,D4=9 D5=44,D6=265,D7=1854,D8=14833 D9=133496,D10=1334961 Calculate the value: D2-2D1, D3-3D2, D4-4D3, D5-5D4, …, D10-10D9 D2−2D1=1−2(0)=1 D4−4D3=9−4(2)=1 D6−6D5=265−6(44)=1 D8−8D7=14833−8(1854)=1 D10−10D9=1334961−10(133496)=1 D3−3D2=2−3(1)=−1 D5−5D4=44−5(9)=−1 D7−7D6=1854−7(265)=−1 D9−9D8=133496−9(14833)=−1 Dn−nDn−1={1 if n is even −1 if n is odd } Thus, the pattern is given as: Dn−nDn−1={1 if n is even −1 if n is odd }