MHB Prove the Following Mathematic Form

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The discussion centers on proving that the expression \( A = \overbrace{ 11-------1 }^{m}\underbrace{ 22-------2 }_{m} \) can be represented as the product of two consecutive integers, \( A = k \times (k + 1) \), where \( k \in \mathbb{N} \). The proof involves manipulating the expression for \( A \) into a form that reveals its factors, ultimately showing that \( A = (10^m - 1)(10^m + 2) / 9 \). This leads to the conclusion that both factors are multiples of 3 and differ by 3, allowing for the division by 9 to yield \( A = k(k + 1) \) with \( k = \frac{10^m - 1}{3} \). The discussion highlights the algebraic steps taken to arrive at this conclusion, confirming the relationship between \( A \) and consecutive integers. The proof is presented as a satisfying mathematical solution.
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$A=\overbrace{ 11-------1 }^{m}\underbrace{ 22-------2 }_{m}$

prove :$A=k\times (k+1),\,\, where\,\, k\in N$

(A can be expressed as the multiplication of two consecutive positive integers)
 
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\displaystyle A=\frac{10^m-1}{9} \times 10^m +2 \times \frac{10^m-1}{9}

\displaystyle 9A=(10^m-1) \times 10^m +2 \times (10^m-1)

\displaystyle 9A=10^{2m} -10^m +2 \times 10^m-2

\displaystyle 9A=10^{2m} +10^m - 2

\displaystyle 9A=(10^m-1)(10^m +2)

Each factor on the right is a multiple of 3 and they differ by 3 so on dividing by 9 we get \displaystyle A=k(k+1) where \displaystyle k = \frac{10^m-1}{3} .

I nice start to my day, thank you :)
 
Hi MR

A nice solution (Clapping)
 
Albert said:
$A=\overbrace{ 11-------1 }^{m}\underbrace{ 22-------2 }_{m}$

prove :$A=k\times (k+1),\,\, where\,\, k\in N$

(A can be expressed as the multiplication of two consecutive positive integers)

$A=\overbrace{ 11-------1 }^{m}\times 10^m+\underbrace{ 1-------1 }_{m}\times 2$
$=\overbrace{ 11-------1 }^{m}\times (10^m+2)=\overbrace{ 11-------1 }^{m}\times (\overbrace{ 99-------9 }^{m}+3)$
$=\overbrace{ 11-------1 }^{m}\times (\overbrace{ 33-------3 }^{m}\times 3+3)$
$=\overbrace{ 33-------3 }^{m}\times (\overbrace{ 33-------3 }^{m}+1)=k\times (k+1)$
($k=\overbrace{ 33-------3 }^{m}$)
 
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