f(n)= 1.Pn + 2.Pn + 3.Pn +...+ N.Pn where Pn = 1/N f(n)= 1.1/

What is the average number of comparisons in a sequential
search?

Question Posted / vivek tiwari

Answer Posted / vivek tiwari

f(n)= 1.Pn + 2.Pn + 3.Pn +...+ N.Pn
where
Pn = 1/N
f(n)= 1.1/N +2.1/N + 3.1/N+....+N.1/N
= (1+2+3+....+N)1/N
= N(N+1)/2N
= (N+1)/2

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47 Yes

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