I'm trying to apply an induction proof to show that $((n+1) 2^n - 1 $ is the sum of $(i 2^{i-1})$ from $0$ to $n$.
- the base case: L.H.S = R.H.S
- we assume that $(k+1) 2^k - 1 $ is true.
- we need to prove that $(k+2) 2^{k+1} - 1$
My try to prove 3 is as follows:
$(k+2) 2^{k+1} - 1$
$(k+2) (2^k * 2) - 1$ , from 2: $2^k = 1/(k+1)$
$(k+2) (2 / (k+1)) - 1$
$(k+1) 2 - 1$
My question is, how could I get to $2^k$ from the last line to prove this formula is right?
Answer
You have a typo in your statement. You want to show $$ \sum_{i=0}^n i 2^{i-1} = n 2^n - 2^n + 1 $$
Base case $n = 0$:
$$ \sum_{i=0}^0 i 2^{i-1} = 0 = 0 - 1 + 1$$
Assume that $$ \sum_{i=0}^n i 2^{i-1} = n 2^n - 2^n + 1$$
Then make an induction step from $n$ to $n+1$. This means you want to show that given your assumption you can show
$$ \sum_{i=0}^{n+1} i 2^{i-1} = (n+1) 2^{n+1} - 2^{n+1} + 1$$
Do this as follows:
$$ \sum_{i=0}^{n+1} i 2^{i-1} = \sum_{i=0}^{n} i 2^{i-1} + (n+1)2^n = n 2^n - 2^n + 1 + (n+1)2^n = n 2^{n+1} + 1 = (n+1)2^{n+1} - 2^{n+1} + 1$$
Which is what you wanted to show. Hope this helps.
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