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Sagot :
To prove that the sum of the series [tex]\(1 + 3 + 6 + \ldots + \frac{n(n+1)}{2} = \frac{n(n+1)(n+2)}{6}\)[/tex] for all natural numbers [tex]\(n\)[/tex], let's follow a step-by-step approach:
### Step 1: Understand the Series
The series is given as [tex]\(1 + 3 + 6 + \ldots + \frac{n(n+1)}{2}\)[/tex]. Notice that each term in the series follows the pattern:
- The [tex]\(k\)[/tex]-th term is given by [tex]\(\frac{k(k+1)}{2}\)[/tex].
### Step 2: Define the Sum
Let [tex]\(S_n\)[/tex] be the sum of the first [tex]\(n\)[/tex] terms of the series:
[tex]\[ S_n = \sum_{k=1}^n \frac{k(k+1)}{2} \][/tex]
We want to prove that:
[tex]\[ S_n = \frac{n(n+1)(n+2)}{6} \][/tex]
### Step 3: Express the Sum Using Known Summation Formulas
The series can be split and simplified using known summation formulas. We rewrite each term [tex]\(\frac{k(k+1)}{2}\)[/tex] as:
[tex]\[ \frac{k(k+1)}{2} = \frac{k^2 + k}{2} \][/tex]
Thus,
[tex]\[ S_n = \sum_{k=1}^n \frac{k^2 + k}{2} = \frac{1}{2} \left( \sum_{k=1}^n k^2 + \sum_{k=1}^n k \right) \][/tex]
### Step 4: Use Summation Formulas
We employ the summation formulas for [tex]\(\sum_{k=1}^n k\)[/tex] and [tex]\(\sum_{k=1}^n k^2\)[/tex]:
[tex]\[ \sum_{k=1}^n k = \frac{n(n+1)}{2} \][/tex]
[tex]\[ \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \][/tex]
### Step 5: Substitute and Simplify
Substitute these formulas back into the expression for [tex]\(S_n\)[/tex]:
[tex]\[ S_n = \frac{1}{2} \left( \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right) \][/tex]
Combine the terms inside the parentheses:
[tex]\[ S_n = \frac{1}{2} \left( \frac{n(n+1)(2n+1) + 3n(n+1)}{6} \right) \][/tex]
[tex]\[ S_n = \frac{1}{2} \left( \frac{n(n+1) (2n+1 + 3)}{6} \right) \][/tex]
[tex]\[ S_n = \frac{1}{2} \left( \frac{n(n+1)(2n+4)}{6} \right) \][/tex]
[tex]\[ S_n = \frac{1}{2} \left( \frac{n(n+1)2(n+2)}{6} \right) \][/tex]
[tex]\[ S_n = \frac{1}{2} \left( \frac{2n(n+1)(n+2)}{6} \right) \][/tex]
[tex]\[ S_n = \frac{n(n+1)(n+2)}{6} \][/tex]
### Conclusion
Thus, we have shown that:
[tex]\[ 1 + 3 + 6 + \ldots + \frac{n(n+1)}{2} = \frac{n(n+1)(n+2)}{6} \][/tex]
This completes the proof for all natural numbers [tex]\(n\)[/tex].
### Step 1: Understand the Series
The series is given as [tex]\(1 + 3 + 6 + \ldots + \frac{n(n+1)}{2}\)[/tex]. Notice that each term in the series follows the pattern:
- The [tex]\(k\)[/tex]-th term is given by [tex]\(\frac{k(k+1)}{2}\)[/tex].
### Step 2: Define the Sum
Let [tex]\(S_n\)[/tex] be the sum of the first [tex]\(n\)[/tex] terms of the series:
[tex]\[ S_n = \sum_{k=1}^n \frac{k(k+1)}{2} \][/tex]
We want to prove that:
[tex]\[ S_n = \frac{n(n+1)(n+2)}{6} \][/tex]
### Step 3: Express the Sum Using Known Summation Formulas
The series can be split and simplified using known summation formulas. We rewrite each term [tex]\(\frac{k(k+1)}{2}\)[/tex] as:
[tex]\[ \frac{k(k+1)}{2} = \frac{k^2 + k}{2} \][/tex]
Thus,
[tex]\[ S_n = \sum_{k=1}^n \frac{k^2 + k}{2} = \frac{1}{2} \left( \sum_{k=1}^n k^2 + \sum_{k=1}^n k \right) \][/tex]
### Step 4: Use Summation Formulas
We employ the summation formulas for [tex]\(\sum_{k=1}^n k\)[/tex] and [tex]\(\sum_{k=1}^n k^2\)[/tex]:
[tex]\[ \sum_{k=1}^n k = \frac{n(n+1)}{2} \][/tex]
[tex]\[ \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \][/tex]
### Step 5: Substitute and Simplify
Substitute these formulas back into the expression for [tex]\(S_n\)[/tex]:
[tex]\[ S_n = \frac{1}{2} \left( \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right) \][/tex]
Combine the terms inside the parentheses:
[tex]\[ S_n = \frac{1}{2} \left( \frac{n(n+1)(2n+1) + 3n(n+1)}{6} \right) \][/tex]
[tex]\[ S_n = \frac{1}{2} \left( \frac{n(n+1) (2n+1 + 3)}{6} \right) \][/tex]
[tex]\[ S_n = \frac{1}{2} \left( \frac{n(n+1)(2n+4)}{6} \right) \][/tex]
[tex]\[ S_n = \frac{1}{2} \left( \frac{n(n+1)2(n+2)}{6} \right) \][/tex]
[tex]\[ S_n = \frac{1}{2} \left( \frac{2n(n+1)(n+2)}{6} \right) \][/tex]
[tex]\[ S_n = \frac{n(n+1)(n+2)}{6} \][/tex]
### Conclusion
Thus, we have shown that:
[tex]\[ 1 + 3 + 6 + \ldots + \frac{n(n+1)}{2} = \frac{n(n+1)(n+2)}{6} \][/tex]
This completes the proof for all natural numbers [tex]\(n\)[/tex].
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