Question 1051728: Prove by mathematical induction
1^2+3^2+5^2+.......upto n terms = n(4n^2-1)/3
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! 1) 1^2+3^2+5^2+.......up to (2n-1)^2 = n(4n^2-1)/3 = n(2n-1)(2n+1)/3
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show that 1) is true for n = 1, that is
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1^2 = 1(3) / 3 = 1
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now assume 1) is true for n, then show it is true for n+1
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n(2n-1)(2n+1)/3 + (2n+1)^2 = (2n+1)(2n^2 -n +6n +3) / 3 =
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(2n+1)(2n^2 +5n +3) /3 = (2n+1)(2n+3)(n+1) / 3
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note that if we substitute (n+1) for n in 1) we get
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(n+1)(2(n+1)-1)(2(n+1)+1)/3 = (n+1)(2n+1)(2n+3)/3
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the statement has been proven with induction
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