Proof by Mathematical Induction:
We are asked to prove that for every natural number $n$,
$$
2 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 2
$$
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.
Step 2: Inductive Hypothesis
Assume that the formula holds for some natural number $k$, i.e.,
$$
2 + 2^2 + 2^3 + \cdots + 2^k = 2^{k+1} - 2
$$
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 $k+1$:
$$
\left(2 + 2^2 + 2^3 + \cdots + 2^k\right) + 2^{k+1}
$$
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.
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 $n$.
Proof by Mathematical Induction:
We are asked to prove that for every natural number $n$,
$$
\frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots + \frac{n}{2^n} = 2 - \frac{n+2}{2^n}
$$
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.
Step 2: Inductive Hypothesis
Assume that the formula holds for some natural number $k$, i.e.,
$$
\frac{1}{2} + \frac{2}{2^2} + \cdots + \frac{k}{2^k} = 2 - \frac{k+2}{2^k}
$$
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:
$$
\left(2 - \frac{k+2}{2^k}\right) + \frac{k+1}{2^{k+1}}
$$
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.
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 $n$.
