Sum of 2ᵏ, or of k/(2ᵏ) for k∈ℕ∩[1,n]

Prove, using induction, that for every natural number n: $${\displaystyle \begin{array}{r}2+2^2+2^3+\cdots +2^n=2^{n+1}-2\end{array}}$$

Proof by Mathematical Induction:

We are asked to prove that for every natural number $n$,

$$2+2^2+2^3+\cdots +2^n=2^{n+1}-2$$

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Step 1: Base Case

Let’s check the base case $n=1$.

Left-hand side (LHS):

$$2=2$$

Right-hand side (RHS):

$$2^{1+1}-2=2^2-2=4-2=2$$

✅ The base case holds.

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Step 2: Inductive Hypothesis

Assume that the formula holds for some natural number $k$, i.e.,

$$2+2^2+2^3+\cdots +2^k=2^{k+1}-2$$

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Step 3: Inductive Step

We need to prove that the formula holds for $k+1$, i.e.,

$$2+2^2+2^3+\cdots +2^k+2^{k+1}=2^{(k+1)+1}-2$$

Start from the left-hand side for $k+1$:

$$(2+2^2+2^3+\cdots +2^k)+2^{k+1}$$

Apply the inductive hypothesis to the first part:

$$(2^{k+1}-2)+2^{k+1}$$

Combine like terms:

$$2^{k+1}+2^{k+1}-2=2\cdot 2^{k+1}-2$$

Simplify the exponent:

$$2^{k+2}-2$$

Check the right-hand side formula for $k+1$:

$$2^{(k+1)+1}-2=2^{k+2}-2$$

✅ Matches exactly.

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Conclusion:

By the principle of mathematical induction, the formula

$$2+2^2+2^3+\cdots +2^n=2^{n+1}-2$$

holds for all natural numbers $n$.

Prove, using induction, that for every natural number n: $${\displaystyle \begin{array}{r}{\displaystyle \frac{1}{2}}+{\displaystyle \frac{2}{2^2}}+{\displaystyle \frac{3}{2^3}}+\cdots +{\displaystyle \frac{n}{2^n}}=2-{\displaystyle \frac{n+2}{2^n}}\end{array}}$$

Proof by Mathematical Induction:

We are asked to prove that for every natural number $n$,

$$\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots +\frac{n}{2^n}=2-\frac{n+2}{2^n}$$

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Step 1: Base Case

Check the base case $n=1$.

Left-hand side (LHS):

$$\frac{1}{2}=0.5$$

Right-hand side (RHS):

$$2-\frac{1+2}{2^1}=2-\frac{3}{2}=2-1.5=0.5$$

✅ The base case holds.

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Step 2: Inductive Hypothesis

Assume that the formula holds for some natural number $k$, i.e.,

$$\frac{1}{2}+\frac{2}{2^2}+\cdots +\frac{k}{2^k}=2-\frac{k+2}{2^k}$$

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Step 3: Inductive Step

Prove that the formula holds for $k+1$, i.e.,

$$\frac{1}{2}+\frac{2}{2^2}+\cdots +\frac{k}{2^k}+\frac{k+1}{2^{k+1}}=2-\frac{(k+1)+2}{2^{k+1}}$$

Start from the left-hand side for $k+1$:

Apply the inductive hypothesis to the first part:

$$(2-\frac{k+2}{2^k})+\frac{k+1}{2^{k+1}}$$

Express $\frac{k+1}{2^{k+1}}$ with denominator $2^k$ to combine terms:

$$2-\frac{k+2}{2^k}+\frac{k+1}{2^{k+1}}$$

Express $\frac{k+1}{2^{k+1}}$ as:

$$\frac{k+1}{2\cdot 2^k}=\frac{k+1}{2^k\cdot 2}=\frac{k+1}{2^{k+1}}$$

It doesn't directly combine, so use a common denominator $2^{k+1}$ for the fractions.

Express $2$ with denominator $2^{k+1}$:

$$2=\frac{2\cdot 2^{k+1}}{2^{k+1}}=\frac{2^{k+2}}{2^{k+1}}$$

Express $\frac{k+2}{2^k}$ with denominator $2^{k+1}$:

$$\frac{k+2}{2^k}=\frac{2(k+2)}{2^{k+1}}$$

Now write the entire left-hand side over denominator $2^{k+1}$:

$$\frac{2^{k+2}-2(k+2)+(k+1)}{2^{k+1}}$$

Compute the numerator:

$$2^{k+2}-2(k+2)+(k+1)$$

Compute step by step:

• $2^{k+2}$ stays as is.
• $-2(k+2)=-2k-4$
• $+(k+1)$

Sum:

$$2^{k+2}-2k-4+k+1=2^{k+2}-k-3$$

The numerator becomes:

$$2^{k+2}-(k+3)$$

Now express the right-hand side formula for $k+1$:

Check if it matches:

$$2-\frac{(k+1)+2}{2^{k+1}}=2-\frac{k+3}{2^{k+1}}=\frac{2\cdot 2^{k+1}}{2^{k+1}}-\frac{k+3}{2^{k+1}}$$

Compute numerator:

$$2\cdot 2^{k+1}=2^{k+2}$$

So numerator:

$$2^{k+2}-(k+3)$$

Indeed, matches exactly.

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Conclusion:

By the principle of mathematical induction, the formula

$$\frac{1}{2}+\frac{2}{2^2}+\cdots +\frac{n}{2^n}=2-\frac{n+2}{2^n}$$

holds for every natural number $n$.