contestada

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

Respuesta :

[tex]\text{Proof by induction:}[/tex]
[tex]\text{Test that the statement holds or n = 1}[/tex]

[tex]LHS = (3 - 2)^{2} = 1[/tex]
[tex]RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS[/tex]
[tex]\text{Thus, the statement holds for the base case.}[/tex]

[tex]\text{Assume the statement holds for some arbitrary term, n= k}[/tex]
[tex]1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}[/tex]

[tex]\text{Prove it is true for n = k + 1}[/tex]
[tex]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}[/tex]

[tex]LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}[/tex]
[tex] = \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}[/tex]
[tex] = \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}[/tex]
[tex] = \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}[/tex]
[tex] = \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}[/tex]
[tex] = \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}[/tex]
[tex] = \frac{k(6k^{2} + 15k + 11) + 2}{}[/tex]
[tex] = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}[/tex]
[tex] = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}[/tex]
[tex] = RHS[/tex]

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

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