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Kombinatorik - Georg Mohr-Konkurrencen

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<strong>Kombinatorik</strong>noter 2012, Kirsten Rosenkilde 14<br />

for alle ikke-negative heltal n < N . For at udnytte induktionsantagelsen benyttes den helt<br />

grundlæggende sammenhæng mellem binomialkoefficienter (sætning 2.1) som siger at m r =<br />

m −1<br />

m −1<br />

<br />

r +<br />

r −1 :<br />

N∑ N + 1 − k<br />

N∑ <br />

(N − 1) + 1 − k<br />

N∑ <br />

(N − 1) + 1 − k<br />

=<br />

+<br />

k<br />

k<br />

k − 1<br />

k =0<br />

k =0<br />

k =0<br />

N∑<br />

−1 N<br />

(N − 1) + 1 − k 0 ∑−1<br />

<br />

(N − 1) + 1 − k N 0<br />

=<br />

+ +<br />

+ +<br />

k<br />

N k − 1 −1<br />

N<br />

k =0<br />

k =1<br />

N∑<br />

−2 <br />

(N − 2) + 1 − k<br />

= F N +<br />

k<br />

k =0<br />

= F N + F N −1 = F N +1 .<br />

Dermed er induktionen fuldført.<br />

Opgave 5.4. Vis at<br />

n∑ <br />

(−1) k −1 n<br />

= 1 + 1 k k 2 + 1 3 + ··· + 1 n<br />

k =1<br />

for alle positive heltal n.<br />

Opgave 5.5. Lad n være et positivt heltal. Vis at<br />

(Hint)<br />

n∑<br />

i =0<br />

1<br />

= n + 1 ∑n+1<br />

2 i<br />

2 n+1 i .<br />

n<br />

i<br />

i =1

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