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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 equal 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!

User Lachtan
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1 Answer

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Hello,


\sum_(i=0)^(k-1)\ (2*i+1)=2*\sum_(i=0)^(k-1)\ i +\sum_(i=0)^(k-1)\ 1\\ = (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
User Anandbibek
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