(2+ 1)(2? +1) = 3+2 of 2? +2 (power of 2)+1(power of 2).
Then multiply 2+~+2+1by the square of 2+6.
In other words, this is the law. If it's a fill-in-the-blank question, that's it.
The answer is the (1+2+4+6+~+2n) power of 2 and the (1+2+4+6+~ 2n-1) power of 2+~+.
If this is a big problem, prove it.
There are many ways to prove it, so just assume that the above idea is correct, which is the result of the first n terms of 2n.
Then the term of n+ 1 is multiplied by another (2 2 (n+1)+1).
The result of the nth term is (1+2+4+6 +~+2n) power+2 (1+2+4+6+~ 2 (n-1) power+~+2.
If your idea is right, then the law of recursion is also right.
There is a quadratic sum of (1+2+4+6+~+2 (n+1)) and a quadratic sum of (1+2+4+6+~+2n+1).
Calculate the power (1+2+4+6+~+2n-1) of (1~+2+0).
Multiplied by (2 2 (n+1)+1) is equal to the power of (1 +2+4+6+~+2 (n+ 1)).
The original formula is proved.