How is $D_n^2=N$ derived from $D_n^2=D_{n-1}^2+1$ considering that $D_1^2=1$ (Feynman Lectures on Physics 6.3)

Question

I don't see how equation 6.8 follows from equation 6.9. Feynman states in his lecture:

"In general, we should expect for $D_{n-1}^2$ its “expected value” $<D_{n-1}^2>$ (by definition!). So $<D_n^2> = <D_{n-1}^2>+1$. We have already shown that $D_1^2=1$; it follows then that $<D_n^2> = N$."

equation 6.8: $<D_n^2> = <D_{n-1}^2>+1$

equation 6.9: $<D_n^2> = N$

You can find the full lecture here: Probability (6.3)

Could someone explain how to approach this derivation?

Answer 1

Let $D_n^2=E_n$, then $E_n=E_{n-1}+1 \implies E_n-E_{n-1}=1 \implies E_n=n+A$ (Aeithimatic Progression: common difference is $1$ and $A$ is first term). One may also do teslescopic summation to get $E_n=n+A$. As $E_1=1$, we get$%A=0$. So $E_n=n \implies D_n=\sqrt{n}.$