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  • one year ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false. 1^2 + 4^2 + 7^2 + ... + (3n-2)^2 = (n(6n^2 - 3n - 1))/2

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  1. anonymous
    • one year ago
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    First check if it's true for \(n=1\) (the base case). \[\frac{1(6\times1^2-3\times1-1)}{2}=\frac{6-3-1}{2}\stackrel{\checkmark}=1\] Assume the statement is true for \(n=k\), i.e. \[1^2+4^2+\cdots+(3k-2)^2=\frac{k(6k^2-3k-1)}{2}\] You want to use this hypothesis to establish that \[1^2+4^2+\cdots+(3k-2)^2+(3(k+1)-2)^2=\frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2}\] From the hypothesis, you know that the first \(k\) terms on the left can be simplified: \[\frac{k(6k^2-3k-1)}{2}+(3(k+1)-2)^2=\frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2}\] Some rewriting of the right side: \[\frac{k(6k^2-3k-1)}{2}+(3(k+1)-2)^2=\frac{(k+1)(6k^2+9k+2)}{2}\] Now it's a matter of establishing that this here is an identity.

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