Feynman's random walk (6.3)

Physics Asked by Benjamin Sauvé on September 1, 2020

In chapter 6 of volume 1 of the Feynman lectures on physics, Feynman’s elaborates on the random walk.

Amidst his lecture, he says this: The expected value of D2N for N>1 can be obtained from DN−1. I simply cannot understand this. If N=1 then D1-1=D0?

He proceeds with this: If, after (N−1) steps, we have DN−1, then after N steps we have DN=DN−1+1 or DN=DN−1−1.

Any help is appreciated greatly.

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