Compare these two formulas: a0+a 1x+a2x2+...+a46x46 and a0+a 1+a2+...+a46.

It is found that A0+a1x+a2x2+... +A46x46 = A0+a1+a2+... x =1+a46.

And A0+a1x+a2x2+...+a46x46 = (x2-x+1) 23 squares.

Substitute 1 into the 23rd square of A0+A 1+A2+ ...+A46 = (1-1) =1.

So A0+a1+A2+...+A46 =1.

To do this kind of problem, we should observe the relationship between binomial expansions and then substitute them into numerical values for calculation.

For example:

If A0-a1+A2-A3+A4-A5+...+A46.

We will find that a0+a 1x+a2x2+...+a46x46 is the same as A0-A 1+A2-A3+A4-A5+ ... when x=- 1 +A46.

So-1 is substituted into the 23rd square of the formula (x2-x+ 1) and the 23rd square of (1+ 1), that is, the 23rd square of 3.