<s>[INST] prove by induction : $\frac{(1-a^n)}{1-a}+\frac{(1-a^n)(1-a^{n-1})}{1-a^2}+...+\frac{(1-a^n)(1-a^{n-1})...(1-a)}{1-a^n}=n$ How to prove the following question by using induction?
$${\frac{(1-a^n)}{1-a}+\frac{(1-a^n)(1-a^{n-1})}{1-a^2}+\frac{(1-a^n)(1-a^{n-1})(1-a^{n-2})}{1-a^3}+...+\frac{(1-a^n)(1-a^{n-1})...(1-a)}{1-a^n}=n}$$
and I am interested in solutions without using Induction. Can anyone show me the steps?
Thanks for your help.
 [/INST] I applied  Andre Nicolas's method. Please see steps below.
$$H_a(n)=\frac{(1-a^n)}{1-a}+\frac{(1-a^n)(1-a^{n-1})}{1-a^2}+\frac{(1-a^n)(1-a^{n-1})(1-a^{n-2})}{1-a^3}+...+\frac{(1-a^n)(1-a^{n-1})...(1-a)}{1-a^n}$$
$$H_a(n+1)=\frac{(1-a^{n+1})}{1-a}+\frac{(1-a^{n+1})(1-a^{n})}{1-a^2}+\frac{(1-a^{n+1})(1-a^{n})(1-a^{n-1})}{1-a^3}+...+\frac{(1-a^{n+1})(1-a^{n})...(1-a)}{1-a^{n+1}}$$
$$H_a(n+1)-H_a(n)=a^n+a^{n-1}(1-a^n)+a^{n-2}(1-a^n)(1-a^{n-1})+...+\frac{(1-a^{n+1})(1-a^{n})...(1-a)}{1-a^{n+1}}$$
$$H_a(n)-H_a(n-1)=a^{n-1}+a^{n-2}(1-a^{n-1})+a^{n-3}(1-a^{n-1})(1-a^{n-2})+...+\frac{(1-a^{n})(1-a^{n-1})...(1-a)}{1-a^{n}}$$
$$(1-a^n)(H_a(n)-H_a(n-1))+a^n=a^n+a^{n-1}(1-a^n)+a^{n-2}(1-a^n)(1-a^{n-1})+a^{n-3}(1-a^n)(1-a^{n-1})(1-a^{n-2})+...+(1-a^n)\frac{(1-a^{n})(1-a^{n-1})...(1-a)}{1-a^{n}}$$
$$(1-a^n)(H_a(n)-H_a(n-1))+a^n=H_a(n+1)-H_a(n) \tag1$$
If we select $H_a(n)=n+c$ (where c is constant),  $H_a(n)=n+c$ satisfies Equation (1).
But we know that $H_a(1)=\frac{(1-a^1)}{1-a}=1$,  so $c$ should be zero
Note: I could not find a method to solve Equation (1) and to find the solution directly  $H_a(n)=n+c$. If someone knows how to prove that I will be happy to see it. 
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