Ytukyg

I'm trying to prove this by mathematical induction, but I just can't seem to get the answer that I should be getting. Here's what I have so far:

Let $P(n)$ be the statement (this is the equation that I'm supposed to prove by induction): $$sum_{i=0}^n (n-i)2^i = 2^{n+1}-n-2,$$
Basis step: $(n=0)$: $$P(0) = (0-0)2^0 = 2^{0+1}-0-2 = 0 = 0.$$
Inductive step:
Assume that $P(n)$ is true, that is, $$sum_{i=0}^n (n-i)2^i = 2^{n+1}-n-2.$$

Showing that $P(n+1)$ is also true, that is:
$$sum_{i=0}^{n+1} (n-i)2^i = 2^{n+2}-(n+1)-2 $$
$$ = 2^{n+2}-n-3 $$
$P(n+1) ={}$ $$sum_{i=0}^{n+1} (n-i)2^i + n-(n+1)2^{n+1}$$
$$ = 2^{n+1}-n-2+(n-n-1)2^{n+1} $$
$$ = 2^{n+1}-n-2-(1)2^{n+1}$$

As can be seen, I am not getting back the result I'm supposed to be getting for $P(n+1)$. Can someone assist me here, please?

Given $$P(color{red}n)=sum_{i=0}^color{red}n (color{red}n-i)2^i,$$ note:
$$begin{align}P(color{red}0)=&sum_{i=0}^color{red}0 (color{red}0-i)2^i = (color{red}0-0)2^0=0=2^{color{red}0+1}-color{red}0-2; text{(base step)}\
P(color{red}1)=&sum_{i=0}^color{red}1 (color{red}1-i)2^i = (color{red}1-0)2^0+(color{red}1-1)2^1=1;\
vdots\
P(color{red}{n})=&sum_{i=0}^{color{red}{n}} (color{red}{n}-i)2^i=color{blue}{2^{n+1}-n-2}; text{(inductive hypothesis)}\
P(color{red}{n+1})=&sum_{i=0}^{color{red}{n+1}} (color{red}{n+1}-i)2^i =\
=&sum_{i=0}^n (n+1-i)2^i+require{cancel} cancel{(n+1-(color{red}{n+1}))2^{n+1}}=\
=&sum_{i=0}^n(n-i)2^i+sum_{i=0}^n2^i=\
=&color{blue}{2^{n+1}-n-2}+2^{n+1}-1=\
=&2^{n+2}-(n+1)-2.end{align}$$

Your $P(n+1)$ is missing a $+1$ (as in, for "$n+1$") inside the sum parenthesis.

Hint:
begin{align}
sum_{i=0}^{n+1} (n+1-i)2^i &=sum_{i=0}^{n+1} (n-i)2^i +sum_{i=0}^{n+1}2^i \
&=(n-(n+1))2^{n+1} +color{blue}{sum_{i=0}^{n} (n-i)2^i} +sum_{i=0}^{n+1}2^i \
end{align}
Now you can use $color{blue}{P(n)}$. Also, you need $displaystyle sum_{i=0}^{n+1} 2^i = 2^{n+2} -1$.