Acoin is tossed until the first time a head turns up. if this occurs on the n th toss and n is odd you win 2n/n , but if n is even then you lose 2n/n . then if your expected winnings exist they are given by the convergent series.

This is a series of variable harmonics that converges to log (2).

Step-by-step explanation:

Let's ignore for a moment that n is even or odd, and look at the expected value for any n. Let X be profit, which can be negative if we lose. The wait is given using (and assuming the coin is valid)

E[X]=∑n=1∞((−1)n+1×2∧n/n)⋅1/2∧n=∑n=1∞(−1)n+1×1/n.

This is a series of variable harmonics that converges to log (2). However, expectation exists only if it absolutely converges! Looking at

∑n=1∞∣(−1)n+1×2∧n/n∣×1/2n=∑n=1∞1/n

we notice that a number of harmonics diverge, therefore, in fact, there is no expectation of X.

arithmetic sequence

step-by-step explanation:

given that a sequence has terms as

19, 15, 11,

this is represented on a graph

the pattern indicates that the sequence is an arithmetic sequence with i term

a=19 and common difference d =-4

any general term

thus the sequence is reduced by 4 for each successive term starting from 1 and nth term is

24-5n.

first negative term would start after 5th term

5th term =24-25< 0

from 5th term onwards the sequence would be negative.

x=6 and x=3

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