# Prove that displaystylesum_{j=1}^{n} 2^{j}=2^{n+1}-2 forallgeq1 Question
Probability and combinatorics Prove that $$\displaystyle\sum_{j=1}^{n} 2^{j}=2^{n+1}-2$$
$$\forall\geq1$$ 2021-01-03
Step 1
To proof: $$\displaystyle\sum_{j=1}^{n} 2^{j}=2^{n+1}-2$$ for all positive integers n.
Proof by induction
Let P(n) be the statement "$$\displaystyle\sum_{j=1}^{n} 2^{j}=2^{n+1}-2$$".
Basis step n=1
$$\displaystyle\sum_{j=1}^{n} 2^{j}=\displaystyle\sum_{j=1}^{1} 2^{j}=2^{1}=2$$
$$2^{n+1}-2=2^{1+1}-2=2^{2}-2=2$$
Thus P(1) is true.
Inductive step
Let P(k) be true
$$\displaystyle\sum_{j=1}^{k} 2^{j}=2^{k+1}-2$$
We need to proof that P(k+1) is true
$$\displaystyle\sum_{j=1}^{k+1} 2^{j}$$
$$=[\displaystyle\sum_{j=1}^{k+1} 2^{j}]+2^{k+1}$$
$$=2^{k+1}-2+2^{k+1} \ \text{Since P(k) is true}$$
$$=2\times2^{k+1}-2 \ \text{Combine loke terms}$$
$$=2^{k+1+1}-2$$
$$=2^{(k+1)+1}-2$$
Thus P(k+1) is true
By the principle if mathematical induction, P(n) is true for all positive integers n.

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