 # Prove that displaystylesum_{j=1}^{n} 2^{j}=2^{n+1}-2 forallgeq1 Brittney Lord 2021-01-02 Answered
Prove that $\sum _{j=1}^{n}{2}^{j}={2}^{n+1}-2$
$\mathrm{\forall }\ge 1$
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Step 1
To proof: $\sum _{j=1}^{n}{2}^{j}={2}^{n+1}-2$ for all positive integers n.
Proof by induction
Let P(n) be the statement "$\sum _{j=1}^{n}{2}^{j}={2}^{n+1}-2$".
Basis step n=1
$\sum _{j=1}^{n}{2}^{j}=\sum _{j=1}^{1}{2}^{j}={2}^{1}=2$
${2}^{n+1}-2={2}^{1+1}-2={2}^{2}-2=2$
Thus P(1) is true.
Inductive step
Let P(k) be true
$\sum _{j=1}^{k}{2}^{j}={2}^{k+1}-2$
We need to proof that P(k+1) is true
$\sum _{j=1}^{k+1}{2}^{j}$
$=\left[\sum _{j=1}^{k+1}{2}^{j}\right]+{2}^{k+1}$

$={2}^{k+1+1}-2$
$={2}^{\left(k+1\right)+1}-2$
Thus P(k+1) is true
By the principle if mathematical induction, P(n) is true for all positive integers n.