Splitting a summation into even and odd components

Question

I'm having trouble understanding how the the second line of the splitting of this summation works. I don't understand how the sums of each individual parts can still run from n = 0 to infinity without over-counting and making the sum larger.

Definition of the matrix's properties

Splitting of summation into even/odd parts (trouble understanding bridge from line 1 to 2)

Thanks for any help.

If I write $2^{2n}+2^{2n+1}$ for $n=0\ldots 4$ you have $$2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9$$ No overcounting as you can see — Raffaele, Sep 26 '17 at 19:42

score 1 · Accepted Answer · answered Sep 26 '17 at 19:39

The reason you are not overcounting is that the exponents in each of the split sums are advancing by $2$ for increase in $n$ or $1$. So if you went up to any particular (even) upper limit of the exponent, you would have the same number of terms in the sum of the split sums and the original sum.

Splitting a summation into even and odd components

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