Compute the unique positive integer n such that 2 \cdot 2^2 + 3 \cdot 2^3 + 4 \cdot 2^4 + \dots + n \cdot 2^n = 32.
2⋅22+3⋅23+4⋅24+⋯+n⋅2n=32.
We can calculate that 2⋅22is 8, and that 3⋅23 is 24, and because 8 + 24 is 32, then n is 3, and the equation 2⋅22+3⋅23+4⋅24+⋯+n⋅2n=32 is 2⋅22+3⋅23=32
Answer: 3