Discrete Mathematics with Graph Theory (3rd Edition) 210

Discrete Mathematics with Graph Theory (3rd Edition) 210 -...

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208 Solutions to Exercises (b) The formula is true if r = n (since then both sides equal 1), so we assume r < n. Then using identity (5) = 1 + [(r + 2) _ (r + 1) + (r + 3) _ (r + 2) r+1 r+1 r+1 r+1 + (r + 4) _ (r + 3) + " . r+1 r+1 as desired. This formula says that any given entry (other than the first 1) in a row of Pascal's triangle is equal to the sum of the entries on the southwest-northeast diagonal (I) beginning with the entry above and to the left of the given one. 20. [BB] Consider (x - y)n = 2:~=o (~)xn-k( _y)k = 2:~=o (~)xn-k( _l)kyk.
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Unformatted text preview: Setting x = y = 1, we obtain 21. (a) 1 + 2(i) + 4@ + 8(~) + 16 = 1 + 2(4) + 4(6) + 8(4) + 16 = 81 = 3 4 . (b) Setting x = 1, Y = 2 in (x + y)n = 2:~=o (~)xn-kyk, we obtain 3 n = (1 + 2t = t (~)ln-k2k = t2k(~). k=O k=O 22. (a) [BB]~k(~) = (~) +2(;) +3(~) +4(~) +&quot;.+(n-1)(n:1) +n n(n -l)(n - 2) n(n -l)(n -2)(n - 3) = n + n(n -1) + + +&quot;. + n(n -1) + n 2 3! = n[l + (n -1) + (n -l)(n - 2) + (n -l)(n -2)(n - 3) +&quot;. + n _ 1 + 1] 2 3! = n [1 + (n ~ 1) + (n ~ 1) + (n; 1) + ... + 1] = n2n-1 using the result of Problem 34 in the last step....
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