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2 answers:
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Hello,
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
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
we have
To find the zeros equate the function to zero
Group terms that contain the same variable, and move the constant to the opposite side of the equation
Factor the leading coefficient
Complete the square. Remember to balance the equation by adding the same constants to each side
Rewrite as perfect squares
square root both sides
therefore
the answer is
The zeros of the quadratic function are
and