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lozanna [386]
3 years ago
13

help 112. if k is a positive integer, then for every value of k, the sum of the k smallest distinct odd positive integers is equ

al to which of the following? f. k g. 2k h. 2k - 1 j. k^2 k. 2k^2 justify your answer, help me understand! thank you!
Mathematics
1 answer:
Ksenya-84 [330]3 years ago
7 0
Hello,

\sum_{i=0}^{k-1}\ (2*i+1)=2*\sum_{i=0}^{k-1}\ i +\sum_{i=0}^{k-1}\ 1\\

= \dfrac{k*(k-1)}{2} +k=k^2-k+k=k^2

In details:

s= 1             +3            +5+7+9+....    +2(k-2)+1      +2(k-1)+1
s= 2(k-1)+1  +2(k-2)+1 +....  +9+7+5  +3                 +1
2s=(2k-2+2) +(2k-4+4)+...........................................+(2k-2+2)
   =2k+2k+2k+....+2k (there is k terms)

2s=2k*k
2s=2k²
s=k²


Answer J
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Step-by-step explanation:

The function given in the question is 6·x² + 48·x + 207 = 15

The intermediate steps in the to express the given function in the form (x + a)² = b are found as follows;

6·x² + 48·x + 207 = 15

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3 years ago
What’s the mean for the data set below? <br> 12, 15, 10, 11
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Learn more about function and types of transformation such as translation, dilation etc here

brainly.com/question/26092237

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