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forsale [732]
3 years ago
12

Illustrate the 7th pattern of the sequence of square numbers. ​

Mathematics
2 answers:
valentinak56 [21]3 years ago
6 0

<em>1</em><em>,</em><em>4</em><em>,</em><em>9</em><em>,</em><em>1</em><em>6</em><em>,</em><em>2</em><em>5</em><em>,</em><em>3</em><em>6</em><em>,</em><em>4</em><em>9</em><em>,</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

<em>7</em><em>th</em><em> </em><em>pattern</em><em> </em><em>=</em><em>4</em><em>9</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

tino4ka555 [31]3 years ago
6 0

Answer:

1, 4, 9, 16, 25, 36, 49…................the 7 the pattern is 49

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Eduardwww [97]

We have 3⁴ = 81, so we can factorize this as a difference of squares twice:

x^8 - \dfrac1{81} = \left(x^2\right)^4 - \left(\dfrac13\right)^4 \\\\ x^8 - \dfrac1{81} = \left(\left(x^2\right)^2 - \left(\dfrac13\right)^2\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)

Depending on the precise definition of "completely" in this context, you can go a bit further and factorize x^2-\frac13 as yet another difference of squares:

x^2 - \dfrac13 = x^2 - \left(\dfrac1{\sqrt3}\right)^2 = \left(x-\dfrac1{\sqrt3}\right)\left(x+\dfrac1{\sqrt3}\right)

And if you're working over the field of complex numbers, you can go even further. For instance,

x^4 + \dfrac19 = \left(x^2\right)^2 - \left(i\dfrac13\right)^2 = \left(x^2 - i\dfrac13\right) \left(x^2 + i\dfrac13\right)

But I think you'd be fine stopping at the first result,

x^8 - \dfrac1{81} = \boxed{\left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)}

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