Question 877695
Recall 1+2+...+n = n(n+1)/2. If we can find a general formula for *[tex \large 1 \cdot 2 + 2 \cdot 3 + 4 \cdot 5 + 8 \cdot 9 + \ldots + 2^n \cdot 2^n + 1], we are essentially done.


Rewrite as *[tex \large (1^2 + 2^2 + 4^2 + \ldots + (2^n)^2) + (1+2+4+ \ldots + 2^n)]. The latter term is equal to *[tex \large 2^{n+1} - 1]. Note that the first sum is geometric with common ratio 4 (can be written as 1^2 + 4^2 + ... + 4^n) and is equal to *[tex \large \frac{4^{n+1} - 1}{3}].


Therefore the sum is *[tex \large \frac{\frac{4^{n+1} - 1}{3} + 2^{n+1} - 1}{2}], which I'll leave you to simplify.