Prove, using induction, that for every natural number n: $$\displaystyle \begin{aligned} 1\cdot 2+2\cdot 3+3\cdot 4+\cdots +n\cdot (n+1)=\dfrac{n(n+1)(n+2)}{3} \end{aligned}$$
Let's prove the statement by mathematical induction.
Statement:
For every natural number $n \geq 1$:
$$ 1\cdot 2 + 2\cdot 3 + 3\cdot 4 + \cdots + n\cdot (n+1) = \frac{n(n+1)(n+2)}{3} $$
Step 1: Base Case
Test for $n = 1$.
Left-hand side (LHS):
$$ 1\cdot 2 = 2 $$
Right-hand side (RHS):
$$ \frac{1(1+1)(1+2)}{3} = \frac{1\cdot 2\cdot 3}{3} = \frac{6}{3} = 2 $$
✅ The base case holds.
Step 2: Inductive Hypothesis
Assume the formula holds for $n = k$, i.e.,
$$ 1\cdot 2 + 2\cdot 3 + \cdots + k\cdot (k+1) = \frac{k(k+1)(k+2)}{3} $$
Step 3: Inductive Step
Show it holds for $n = k + 1$.
Left-hand side for $n = k + 1$:
$$ \left(1\cdot 2 + 2\cdot 3 + \cdots + k\cdot (k+1)\right) + (k+1)(k+2) $$
Using the inductive hypothesis:
$$ \frac{k(k+1)(k+2)}{3} + (k+1)(k+2) $$
Factor out $(k+1)(k+2)$:
$$ (k+1)(k+2)\left( \frac{k}{3} + 1 \right) $$
Combine the terms inside the parentheses:
$$ (k+1)(k+2) \cdot \frac{k+3}{3} $$
Rearrange:
$$ \frac{(k+1)(k+2)(k+3)}{3} $$
This matches the RHS for $n = k+1$.
✅ Conclusion:
By mathematical induction, the formula holds for all natural numbers $n \geq 1$:
$$ 1\cdot 2 + 2\cdot 3 + \cdots + n\cdot (n+1) = \frac{n(n+1)(n+2)}{3} $$
Prove, using induction, that for every natural number n: $$\displaystyle \begin{aligned} \dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\cdots +\dfrac{1}{n\cdot (n+1)}=\dfrac{n}{n+1} \end{aligned}$$
Let's prove this formula using mathematical induction.
Statement:
For every natural number $n \geq 1$:
$$ \frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{1}{n\cdot (n+1)} = \frac{n}{n+1} $$
Step 1: Base Case
Test for $n = 1$.
Left-hand side (LHS):
$$ \frac{1}{1\cdot 2} = \frac{1}{2} $$
Right-hand side (RHS):
$$ \frac{1}{1+1} = \frac{1}{2} $$
✅ The base case holds.
Step 2: Inductive Hypothesis
Assume the formula holds for $n = k$, i.e.,
$$ \frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{1}{k\cdot (k+1)} = \frac{k}{k+1} $$
Step 3: Inductive Step
Show it holds for $n = k + 1$.
Left-hand side for $n = k + 1$:
$$ \left(\frac{1}{1\cdot 2} + \cdots + \frac{1}{k\cdot (k+1)}\right) + \frac{1}{(k+1)(k+2)} $$
Apply the inductive hypothesis:
$$ \frac{k}{k+1} + \frac{1}{(k+1)(k+2)} $$
Get a common denominator:
$$ \frac{k(k+2) + 1}{(k+1)(k+2)} $$
Compute the numerator:
$$ k(k+2) + 1 = k^2 + 2k + 1 = (k+1)^2 $$
So the sum becomes:
$$ \frac{(k+1)^2}{(k+1)(k+2)} $$
Cancel one $(k+1)$ from numerator and denominator:
$$ \frac{k+1}{k+2} $$
This matches the RHS for $n = k + 1$:
$$ \frac{k+1}{(k+1)+1} = \frac{k+1}{k+2} $$
✅ Conclusion:
By mathematical induction, the formula holds for all natural numbers $n \geq 1$:
$$ \frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{1}{n\cdot (n+1)} = \frac{n}{n+1} $$
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