Mathematics, 21.12.2019 02:31 igtguith
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
Answers: 3
![RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}](/tpl/images/0428/5630/32ffc.png)
![LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}](/tpl/images/0428/5630/3475c.png)
![= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}](/tpl/images/0428/5630/47b90.png)
![= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}](/tpl/images/0428/5630/57ae5.png)
![= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}](/tpl/images/0428/5630/e416a.png)
![= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}](/tpl/images/0428/5630/0b5ec.png)
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Use mathematical induction to prove the statement is true for all positive integers n, or show why i...
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