Expert answers to "Prove that ∑j=1n2j=2n+1−2 ∀≥1"

Brittney Lord

Answered question

2021-01-02

Prove that

\sum_{j = 1}^{n} 2^{j} = 2^{n + 1} - 2

\forall \geq 1

cheekabooy

Skilled2021-01-03Added 83 answers

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}

= [\sum_{j = 1}^{k + 1} 2^{j}] + 2^{k + 1}

= 2^{k + 1} - 2 + 2^{k + 1} Since P(k) is true

= 2 \times 2^{k + 1} - 2 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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