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base case:
1
j=1
4(1)+3 = 7
2(1)^2 + 5(1) = 7
Inductive step: Assume that for any k >= 1,
k
j=1
(4j+3) = 2k^2 + 5k
We will prove that
k+1
j=1
(4j+3) = 2(k+1)^2 + 5(k+1)
k+1
j=1
4j+3 =
k
j=1
4j+3 + 4k + 7
2k^2 + 9k + 7 by inductive hypothesis
(2k^2 + 4k + 2) + 5(k+1) 2(k^2 + 2k+1) + 5(k+1) 2(k+1)^2+5(k+1)
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