SOLUTION:
Step 1:
In this question, we are meant to make a conjecture about the sum of the first 41 positive odd numbers:
[tex]1,\text{ 3, 5 , 7, 9, 11, }\ldots.[/tex][tex]\begin{gathered} \text{Using the sum of the first 41 terms ( Arithmetic progression)} \\ S_{\text{n }}=\text{ }\frac{n}{2}\lbrack\text{ 2a + ( n - 1) d\rbrack} \\ Sn=\text{ Sum of n terms} \\ n\text{ = number of terms = 41} \\ \text{a = first term = 1} \\ d\text{ = common diference = 3 - 1 = 2} \end{gathered}[/tex]Step 2:
[tex]\begin{gathered} S_{41}=\text{ }\frac{41}{2}\lbrack\text{ 2(1) + ( 41- 1) 2\rbrack} \\ S_{41}=\text{ }\frac{41}{2}\lbrack\text{ 2 + ( 40 x 2 ) \rbrack} \\ S_{41}=\text{ }\frac{41}{2}\lbrack\text{ 2 + 80\rbrack} \\ S_{41}=\text{ }\frac{41}{2}\text{ X }\frac{82}{1} \\ S_{41}\text{ = 41 X 41 = 1681} \end{gathered}[/tex]Step 3:
Part 1: From the above calculations, we can see clearly that:
This corresponds to OPTION 2 :
The sum is equal to the number of terms squared.
Part 2: The sum in a whole number is 1681