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Sagot :
To solve this problem, we need to find the value of [tex]\(n\)[/tex] such that the arithmetic mean (AM) between [tex]\(a\)[/tex] and [tex]\(b\)[/tex] is equal to [tex]\(\frac{a^{n+1} + b^{n+1}}{a^n + b^n}\)[/tex].
The arithmetic mean of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] is given by:
[tex]\[ \text{AM} = \frac{a + b}{2} \][/tex]
According to the problem, this is equal to [tex]\(\frac{a^{n+1} + b^{n+1}}{a^n + b^n}\)[/tex]. Therefore, we set up the following equation:
[tex]\[ \frac{a + b}{2} = \frac{a^{n+1} + b^{n+1}}{a^n + b^n} \][/tex]
To eliminate the fractions, we cross-multiply:
[tex]\[ (a + b)(a^n + b^n) = 2(a^{n+1} + b^{n+1}) \][/tex]
Now, let's expand and simplify the left side:
[tex]\[ a^{n+1} + ab^n + ba^n + b^{n+1} = 2a^{n+1} + 2b^{n+1} \][/tex]
Rearranging all terms to one side gives:
[tex]\[ a^{n+1} + b a^n + a b^n + b^{n+1} - 2a^{n+1} - 2b^{n+1} = 0 \][/tex]
Grouping like terms, we get:
[tex]\[ -a^{n+1} + a b^n + b a^n - b^{n+1} = 0 \][/tex]
We can rewrite this as:
[tex]\[ a b^n + b a^n - a^{n+1} - b^{n+1} = 0 \][/tex]
Factor out common terms:
[tex]\[ a (b^n - a^n) + b (a^n - b^n) = 0 \][/tex]
Notice that [tex]\(b^n - a^n\)[/tex] and [tex]\(a^n - b^n\)[/tex] are opposites, so combining these terms gives:
[tex]\[ a b^n - a^{n+1} + b a^n - b^{n+1} = 0 \][/tex]
Which simplifies to:
[tex]\[ (ab^n - a^{n+1}) + (ba^n - b^{n+1}) = 0 \][/tex]
or:
[tex]\[ -a^{n+1} + a b^n + b a^n - b^{n+1} = 0 \][/tex]
Examining this expression, it must be true for all [tex]\(a\)[/tex] and [tex]\(b\)[/tex]. To satisfy the equation identically, consider the case where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are equal. If we let [tex]\(a = b\)[/tex], then:
[tex]\[ a(a^n - a^n) + b(a^n - a^n) = 0 \][/tex]
This simplifies to:
[tex]\[ 0 = 0 \][/tex]
Independently of [tex]\(a\)[/tex] and [tex]\(b\)[/tex], this simplification holds valid. To find [tex]\(n\)[/tex], let's explore if a general [tex]\(n = 1\)[/tex] would simplify the problem. Substituting [tex]\(n = 1\)[/tex], let's check:
[tex]\[ \frac{a + b}{2} = \frac{a^{1+1} + b^{1+1}}{a^1 + b^1} = \frac{a^2 + b^2}{a + b} \][/tex]
This implies:
[tex]\[ \frac{a + b}{2} = \frac{a^2 + b^2}{a + b} \][/tex]
Multiplying both sides by [tex]\(a + b\)[/tex] leads to:
[tex]\[ a + b = \frac{a^2 + b^2}{a + b} \][/tex]
Thus, simplifying:
[tex]\[ a + b = \frac{a^2 + b^2}{a + b} \][/tex]
Upon rearrangement and ensuring the correctness:
[tex]\[ a + b = a + b \][/tex]
Therefore, the value of [tex]\(n\)[/tex] that satisfies this equation is indeed [tex]\( \boxed{1} \)[/tex].
The arithmetic mean of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] is given by:
[tex]\[ \text{AM} = \frac{a + b}{2} \][/tex]
According to the problem, this is equal to [tex]\(\frac{a^{n+1} + b^{n+1}}{a^n + b^n}\)[/tex]. Therefore, we set up the following equation:
[tex]\[ \frac{a + b}{2} = \frac{a^{n+1} + b^{n+1}}{a^n + b^n} \][/tex]
To eliminate the fractions, we cross-multiply:
[tex]\[ (a + b)(a^n + b^n) = 2(a^{n+1} + b^{n+1}) \][/tex]
Now, let's expand and simplify the left side:
[tex]\[ a^{n+1} + ab^n + ba^n + b^{n+1} = 2a^{n+1} + 2b^{n+1} \][/tex]
Rearranging all terms to one side gives:
[tex]\[ a^{n+1} + b a^n + a b^n + b^{n+1} - 2a^{n+1} - 2b^{n+1} = 0 \][/tex]
Grouping like terms, we get:
[tex]\[ -a^{n+1} + a b^n + b a^n - b^{n+1} = 0 \][/tex]
We can rewrite this as:
[tex]\[ a b^n + b a^n - a^{n+1} - b^{n+1} = 0 \][/tex]
Factor out common terms:
[tex]\[ a (b^n - a^n) + b (a^n - b^n) = 0 \][/tex]
Notice that [tex]\(b^n - a^n\)[/tex] and [tex]\(a^n - b^n\)[/tex] are opposites, so combining these terms gives:
[tex]\[ a b^n - a^{n+1} + b a^n - b^{n+1} = 0 \][/tex]
Which simplifies to:
[tex]\[ (ab^n - a^{n+1}) + (ba^n - b^{n+1}) = 0 \][/tex]
or:
[tex]\[ -a^{n+1} + a b^n + b a^n - b^{n+1} = 0 \][/tex]
Examining this expression, it must be true for all [tex]\(a\)[/tex] and [tex]\(b\)[/tex]. To satisfy the equation identically, consider the case where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are equal. If we let [tex]\(a = b\)[/tex], then:
[tex]\[ a(a^n - a^n) + b(a^n - a^n) = 0 \][/tex]
This simplifies to:
[tex]\[ 0 = 0 \][/tex]
Independently of [tex]\(a\)[/tex] and [tex]\(b\)[/tex], this simplification holds valid. To find [tex]\(n\)[/tex], let's explore if a general [tex]\(n = 1\)[/tex] would simplify the problem. Substituting [tex]\(n = 1\)[/tex], let's check:
[tex]\[ \frac{a + b}{2} = \frac{a^{1+1} + b^{1+1}}{a^1 + b^1} = \frac{a^2 + b^2}{a + b} \][/tex]
This implies:
[tex]\[ \frac{a + b}{2} = \frac{a^2 + b^2}{a + b} \][/tex]
Multiplying both sides by [tex]\(a + b\)[/tex] leads to:
[tex]\[ a + b = \frac{a^2 + b^2}{a + b} \][/tex]
Thus, simplifying:
[tex]\[ a + b = \frac{a^2 + b^2}{a + b} \][/tex]
Upon rearrangement and ensuring the correctness:
[tex]\[ a + b = a + b \][/tex]
Therefore, the value of [tex]\(n\)[/tex] that satisfies this equation is indeed [tex]\( \boxed{1} \)[/tex].
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