Prove that displaystylesum_{j=1}^{n} 2^{j}=2^{n+1}-2 forallgeq1

Question
Prove that \(\displaystyle\sum_{j=1}^{n} 2^{j}=2^{n+1}-2\)
\(\forall\geq1\)

Answers (1)

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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