Sum of products of consecutive naturals, and sum of the reciprocals of such products

Prove, using induction, that for every natural number n: $${\displaystyle \begin{array}{r}1\cdot 2+2\cdot 3+3\cdot 4+\cdots +n\cdot (n+1)={\displaystyle \frac{n(n+1)(n+2)}{3}}\end{array}}$$

Let's prove the statement by mathematical induction.

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Statement:

For every natural number $n\ge 1$:

$$1\cdot 2+2\cdot 3+3\cdot 4+\cdots +n\cdot (n+1)=\frac{n(n+1)(n+2)}{3}$$

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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.

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Step 2: Inductive Hypothesis

Assume the formula holds for $n=k$, i.e.,

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

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Step 3: Inductive Step

Show it holds for $n=k+1$.

Left-hand side for $n=k+1$:

$$(1\cdot 2+2\cdot 3+\cdots +k\cdot (k+1))+(k+1)(k+2)$$

Using the inductive hypothesis:

$$\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$$

Factor out $(k+1)(k+2)$:

$$(k+1)(k+2)(\frac{k}{3}+1)$$

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 $n=k+1$.

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✅ Conclusion:

By mathematical induction, the formula holds for all natural numbers $n\ge 1$:

$$1\cdot 2+2\cdot 3+\cdots +n\cdot (n+1)=\frac{n(n+1)(n+2)}{3}$$

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

Let's prove this formula using mathematical induction.

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Statement:

For every natural number $n\ge 1$:

$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots +\frac{1}{n\cdot (n+1)}=\frac{n}{n+1}$$

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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.

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Step 2: Inductive Hypothesis

Assume the formula holds for $n=k$, i.e.,

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

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Step 3: Inductive Step

Show it holds for $n=k+1$.

Left-hand side for $n=k+1$:

$$(\frac{1}{1\cdot 2}+\cdots +\frac{1}{k\cdot (k+1)})+\frac{1}{(k+1)(k+2)}$$

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 $n=k+1$:

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

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✅ Conclusion:

By mathematical induction, the formula holds for all natural numbers $n\ge 1$:

$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots +\frac{1}{n\cdot (n+1)}=\frac{n}{n+1}$$