```
1
2, 2
3, 4, 3
4, 6, 6, 4
5, 8, 9, 8, 5
6,10, 12, 12, 10, 6
```

The first term just increases by 1 from the term diagonal above and right of it, the second term increases by 2, the third term increases by 3 from what’s diagonally above it etc.

To create the final coefficients the first term is multiplied by 1, the second term by 2, the third term by 3, etc.

The mean:

In[6]:= Sum[ (i*kb^(N – i – 1)*kf^(i – 1))/kf^(N – 1), {i, N – 1}]

This sum can be re-written as:

Out[6]= -((

kf^-N (-kb^N kf + kf^(1 + N) + kb kf^N N – kf^(1 + N) N))/(kb – kf)^2)

The coefficient of variation all follow this pattern:

(derived by examining the triangle above).

In[8]:= eta2 =

1 – kf^(N – 1)*

Sum[ (i + 1)*(i + i*(N – 2 – i))*kb^(N – 2 – i)*kf^(i – 1), {i,

N – 2}]/(Sum[ (i*kb^(N – i – 1)*kf^(i – 1)), {i, N – 1}])^2

Out[8]= 1 – (kf^

N (-4 kb^(1 + N) kf – 2 kb^N kf^2 + 4 kb kf^(1 + N) +

2 kf^(2 + N) + 2 kb^(1 + N) kf N – 2 kb^N kf^2 N +

kb^2 kf^N N + 2 kb kf^(1 + N) N – 3 kf^(2 + N) N +

kb^2 kf^N N^2 – 2 kb kf^(1 + N) N^2 + kf^(2 + N) N^2))/(-kb^

N kf + kf^(1 + N) + kb kf^N N – kf^(1 + N) N)^2

In[9]:= FullSimplify (eta2)

Out[9]= FullSimplify (1 – (kf^

N (-4 kb^(1 + N) kf – 2 kb^N kf^2 + 4 kb kf^(1 + N) +

2 kf^(2 + N) + 2 kb^(1 + N) kf N – 2 kb^N kf^2 N +

kb^2 kf^N N + 2 kb kf^(1 + N) N – 3 kf^(2 + N) N +

kb^2 kf^N N^2 – 2 kb kf^(1 + N) N^2 + kf^(2 + N) N^2))/(-kb^

N kf + kf^(1 + N) + kb kf^N N – kf^(1 + N) N)^2)

In[26]:= eta2 /. {kb -> 5, kf -> 1, N -> 4}

Out[26]= 709/722

eta2B = (N – 4 K – (N + 1) K^2 + 4 (N – N K + 1) K^(N + 1) +

K^(2 N + 2))/((N – N K + K (K^N – 1))^2)

Out[12]= (-4 K + K^(2 + 2 N) + N – K^2 (1 + N) +

4 K^(1 + N) (1 + N – K N))/(K (-1 + K^N) + N – K N)^2

In[20]:= eta2B /. {K -> 5, N -> 3}

Out[20]= 709/722