Question

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!

Answer 1

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