Prove that the following statement holds for all natural numbers:
$$1\cdot2\cdot2^{n} + 2\cdot3\cdot2^{n-1} + \dots +n\cdot(n+1)\cdot2+(n+1)\cdot(n+2) = 2^{n+4}-(n^2-7n+14)$$
I don't need help with the whole proof, just with the base step I somehow managed not to do the right way.
So I recognized that the statement on the left side is a sum of products that have three factors (first two gradually increasing, the third one decreasing). I noted that the following way:
$$\sum_{i=1}^{n+1}i(i+1)2^{n+1-i} = 2^{n+4} - (n^2-7n+14)$$
$$\sum_{i=0}^{n}(i+1)(i+2)2^{n-i}= 2^{n+4} - (n^2-7n+14)$$
Establishing the base:
for $n=1$ we have:
$$(0+1)(0+2)2^{1-0} + (1+1)(1+2)2^{1-1} = 2^{1+4} - (1-7+14)$$ $$1\cdot2\cdot2 + 2\cdot3\cdot2^0= 2^{5} -12$$ $$4+6=2^5-12$$ $$10\neq20$$
I am aware that the base case is not always $n=1$, however, in the solution the base was established for $n=1$ after dividing everything with $2^n$. I wanted to solve this my way first and got stuck right at the beginning.
Thank you in advance.
(1) In your calculation, $(1-7+14)$ should be $8$ and not $12$. But this does not really matter.
(2) I think there is a mistake in the equation, the right side should be $ 2^{n+4}-(n^2+7n+14)$ and everything will work.
A simple test for $n=2:$ $2^6-(4+14+14)=32=1\cdot2\cdot4+2\cdot3\cdot2+3\cdot4$