$\displaystyle \sum_{i=1}^{n+1}i\cdot2^i=2^{n+2}+2$ for all $\displaystyle n\geq 0$.
Hello everyone
Thanks to Renji Rodrigo for this:Adapting this formula to look more like the original, we get
$\displaystyle \sum^{n+1}_{i=1} i\cdot2^{i} = 2^{n+2}n+2$
So let $\displaystyle P(n)$ be the propositional function $\displaystyle S_n=\sum^{n+1}_{i=1} i\cdot2^{i} = 2^{n+2}n+2$
$\displaystyle P(1)$ is $\displaystyle S_1=\sum^{2}_{i=1} i\cdot2^{i} = 2^{3}\cdot1+2$
i.e. $\displaystyle S_1=1\cdot2 + 2\cdot2^2 = 10 = 2^{3}\cdot1+2$
So $\displaystyle P(1)$ is true.
Now $\displaystyle P(n) \Rightarrow S_{n+1} = S_n+(n+2)\cdot2^{n+2}$
$\displaystyle =2^{n+2}n+2+(n+2)\cdot2^{n+2}$
$\displaystyle =2^{n+2}(n+n+2)+2$
$\displaystyle =2^{n+2}(2n+2)+2$
$\displaystyle =2^{n+2}\cdot2(n+1)+2$
$\displaystyle =2^{(n+1)+2}\cdot(n+1)+2$
i.e. $\displaystyle P(n) \Rightarrow P(n+1)$
So $\displaystyle P(n)$ is true for all $\displaystyle n \ge 1$
Grandad