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Sagot :
Sure, let's prove the given summation formula step-by-step.
We need to prove that:
[tex]\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+4)}{6} \][/tex]
The well-known result for the sum of squares formula is:
[tex]\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \][/tex]
However, the formula given in the problem is different. Before proving, let's compare them. The given formula is:
[tex]\[ \frac{n(n+1)(2n + 4)}{6} \][/tex]
Firstly, we'll simplify the given formula and see if it matches the well-known result:
Step 1: Simplify the Given Formula
[tex]\[ \frac{n(n+1)(2n+4)}{6} = \frac{n(n+1)(2n+4)}{6} \][/tex]
Factor out a common term in the numerator:
[tex]\[ = \frac{n(n+1)2(n+2)}{6} \][/tex]
[tex]\[ = \frac{2n(n+1)(n+2)}{6} \][/tex]
Simplify by dividing by 2:
[tex]\[ = \frac{n(n+1)(n+2)}{3} \][/tex]
Step 2: Compare with the Known Formula
We know that:
[tex]\[ \frac{n(n+1)(2n+1)}{6} \][/tex]
It's clear now that:
[tex]\[ \frac{n(n+1)(2n+1)}{6} \][/tex] and [tex]\[ \frac{n(n+1)(n+2)}{3} \][/tex]
are not the same, so it appears there was a typo in the given formula of the problem.
Given this correction, the correct formula for the sum of squares is:
[tex]\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \][/tex]
---
Proof of Sum of Squares Formula
Let's prove the correct formula for the sum of squares [tex]\(\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\)[/tex].
1. Inductive Step:
- Base Case: [tex]\(n = 1\)[/tex]
[tex]\[ \sum_{i=1}^{1} i^2 = 1^2 = 1 \][/tex]
[tex]\[ \frac{1(1+1)(2 \cdot 1 + 1)}{6} = \frac{1 \cdot 2 \cdot 3}{6} = 1 \][/tex]
Hence, the base case holds true.
2. Inductive Hypothesis:
Assume the formula holds for [tex]\(n = k\)[/tex].
[tex]\[ \sum_{i=1}^{k} i^2 = \frac{k(k+1)(2k+1)}{6} \][/tex]
3. Inductive Step:
We need to prove it for [tex]\(n = k+1\)[/tex].
[tex]\[ \sum_{i=1}^{k+1} i^2 = \sum_{i=1}^{k} i^2 + (k+1)^2 \][/tex]
Using the inductive hypothesis:
[tex]\[ \sum_{i=1}^{k+1} i^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2 \][/tex]
Factor out [tex]\((k + 1)\)[/tex]:
[tex]\[ \frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6} \][/tex]
[tex]\[ = \frac{(k+1) [k(2k+1) + 6(k+1)]}{6} \][/tex]
Simplify inside the bracket:
[tex]\[ = \frac{(k+1) [2k^2 + k + 6k + 6]}{6} \][/tex]
[tex]\[ = \frac{(k+1)(2k^2 + 7k + 6)}{6} \][/tex]
Factor out the quadratic inside:
[tex]\[ = \frac{(k+1)(k + 2)(2k + 3)}{6} \][/tex]
Therefore, substituting [tex]\(k + 1\)[/tex] for [tex]\(n\)[/tex] shows that it matches the known formula:
[tex]\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \][/tex]
This completes the proof. The given formula appears to be incorrect, but we've proven the correct formula for the sum of squares.
We need to prove that:
[tex]\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+4)}{6} \][/tex]
The well-known result for the sum of squares formula is:
[tex]\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \][/tex]
However, the formula given in the problem is different. Before proving, let's compare them. The given formula is:
[tex]\[ \frac{n(n+1)(2n + 4)}{6} \][/tex]
Firstly, we'll simplify the given formula and see if it matches the well-known result:
Step 1: Simplify the Given Formula
[tex]\[ \frac{n(n+1)(2n+4)}{6} = \frac{n(n+1)(2n+4)}{6} \][/tex]
Factor out a common term in the numerator:
[tex]\[ = \frac{n(n+1)2(n+2)}{6} \][/tex]
[tex]\[ = \frac{2n(n+1)(n+2)}{6} \][/tex]
Simplify by dividing by 2:
[tex]\[ = \frac{n(n+1)(n+2)}{3} \][/tex]
Step 2: Compare with the Known Formula
We know that:
[tex]\[ \frac{n(n+1)(2n+1)}{6} \][/tex]
It's clear now that:
[tex]\[ \frac{n(n+1)(2n+1)}{6} \][/tex] and [tex]\[ \frac{n(n+1)(n+2)}{3} \][/tex]
are not the same, so it appears there was a typo in the given formula of the problem.
Given this correction, the correct formula for the sum of squares is:
[tex]\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \][/tex]
---
Proof of Sum of Squares Formula
Let's prove the correct formula for the sum of squares [tex]\(\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\)[/tex].
1. Inductive Step:
- Base Case: [tex]\(n = 1\)[/tex]
[tex]\[ \sum_{i=1}^{1} i^2 = 1^2 = 1 \][/tex]
[tex]\[ \frac{1(1+1)(2 \cdot 1 + 1)}{6} = \frac{1 \cdot 2 \cdot 3}{6} = 1 \][/tex]
Hence, the base case holds true.
2. Inductive Hypothesis:
Assume the formula holds for [tex]\(n = k\)[/tex].
[tex]\[ \sum_{i=1}^{k} i^2 = \frac{k(k+1)(2k+1)}{6} \][/tex]
3. Inductive Step:
We need to prove it for [tex]\(n = k+1\)[/tex].
[tex]\[ \sum_{i=1}^{k+1} i^2 = \sum_{i=1}^{k} i^2 + (k+1)^2 \][/tex]
Using the inductive hypothesis:
[tex]\[ \sum_{i=1}^{k+1} i^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2 \][/tex]
Factor out [tex]\((k + 1)\)[/tex]:
[tex]\[ \frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6} \][/tex]
[tex]\[ = \frac{(k+1) [k(2k+1) + 6(k+1)]}{6} \][/tex]
Simplify inside the bracket:
[tex]\[ = \frac{(k+1) [2k^2 + k + 6k + 6]}{6} \][/tex]
[tex]\[ = \frac{(k+1)(2k^2 + 7k + 6)}{6} \][/tex]
Factor out the quadratic inside:
[tex]\[ = \frac{(k+1)(k + 2)(2k + 3)}{6} \][/tex]
Therefore, substituting [tex]\(k + 1\)[/tex] for [tex]\(n\)[/tex] shows that it matches the known formula:
[tex]\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \][/tex]
This completes the proof. The given formula appears to be incorrect, but we've proven the correct formula for the sum of squares.
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