Deduce a formula (in terms of n) for the following sum: sum (2^i * i) where 1<=i<=n, n,i: natural numbers ( one can write this sum as: 1*2^1+ 2*2^2+ .. +n*2^n)

Let S = sum i* 2ⁱ (i=1,2..n) We multiply this by 2 and we obtain:2S = sum (i2ⁱ⁺¹ ) = sum [(i+1-1) * 2ⁱ⁺¹ ]= sum[ (i+1)2ⁱ⁺¹ ] - sum 2ⁱ⁺¹ i=1,2..n ()
sum (i+1)2ⁱ⁺¹ = 22² + .. n2ⁿ + (n+1)* 2ⁿ⁺¹ = S + (n+1) * 2ⁿ⁺¹ - 1*2¹ = S + (n+1)2ⁿ⁺¹ -2
Note that the second sum is the sum of a geometric series of ratio 2, it is equal to 2ⁿ⁺² - 4. Here is a quick proof:Let Z = sum 2ⁱ⁺¹ = 2² + 2³+ .. + 2ⁿ⁺¹ . 2Z = sum 2ⁱ⁺² = 2³ + .. + 2ⁿ⁺¹ + 2ⁿ⁺² . So, 2Z-Z = Z = 2ⁿ⁺² - 2²
Going back to () we obtain that 2S =S + (n+1) * 2ⁿ⁺¹ - 2 - (2ⁿ⁺² - 4), which implies that S = (n+1) * 2ⁿ⁺¹ - 2ⁿ⁺² + 2