Answered

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20. Look at the table. Make a conjecture about the sum of the first 10 positive even numbers.

[tex]\[
\begin{array}{|c|c|}
\hline
2 & = 2 = 1 \cdot 2 \\
\hline
2 + 4 & = 6 = 2 \cdot 3 \\
\hline
2 + 4 + 6 & = 12 = 3 \cdot 4 \\
\hline
2 + 4 + 6 + 8 & = 20 = 4 \cdot 5 \\
\hline
2 + 4 + 6 + 8 + 10 & = 30 = 5 \cdot 6 \\
\hline
\end{array}
\][/tex]

Sagot :

GinnyAnswer
Sure! Let's take a closer look at the pattern in the table of sums of the first few positive even numbers:

1. [tex]\( 2 = 1 \times 2 \)[/tex]
2. [tex]\( 2 + 4 = 6 = 2 \times 3 \)[/tex]
3. [tex]\( 2 + 4 + 6 = 12 = 3 \times 4 \)[/tex]
4. [tex]\( 2 + 4 + 6 + 8 = 20 = 4 \times 5 \)[/tex]
5. [tex]\( 2 + 4 + 6 + 8 + 10 = 30 = 5 \times 6 \)[/tex]

From the table, we notice that the sum of the first [tex]\( n \)[/tex] positive even numbers [tex]\( 2 + 4 + 6 + \ldots \)[/tex] can be represented as [tex]\( n \times (n + 1) \)[/tex].

To verify this conjecture for [tex]\( n = 10 \)[/tex], we can substitute [tex]\( n = 10 \)[/tex] into the formula:

[tex]\[ 10 \times (10 + 1) \][/tex]

Calculating this gives:

[tex]\[ 10 \times 11 = 110 \][/tex]

Thus, the sum of the first 10 positive even numbers is 110. This confirms that our conjecture matches the pattern observed in the table. The sum of the first 10 positive even numbers is indeed 110.
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