Answer
Basic step:
P(0) is true because 12 = 1 = (0+1)(2·0+1)(2·0+3)/3.
Work Step by Step
-Let
P(n) be “12+32+· · ·+(2n+1)2 =(n+1)(2n+1)(2n+3)/3.”
Basic step:
P(0) is true because 12 = 1 = (0+1)(2·0+1)(2·0+3)/3.
--Inductive step:
Assume
-that P(k) is true.
Then
12+32+· · ·+(2k+1)2 +[2(k+1)+1]2
= (k+1)(2k+1)(2k+3)/3+(2k+3)2
= (2k+3)[(k+1)(2k+1)/3+(2k+3)]
= (2k+3)(2k2+9k+10)/3 = (2k+3)(2k+5)(k+2)/3
= [(k+1)+1][2(k+1)+1][2(k+1)+3]/3.
