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Prove that for any positive integer n, the following sum is a perfect square:
$1+8+16+...+8n$.
I have tried to solve the problem by trying to find the equation that would give me the desired values, but since the values differ by so much with each increment on $n$, I think the expression would be exponential. However, all the equations I have tried stop working after some number $n$. What is the equation, prove it using induction, and how should I approach future problems like this asking me to find a term with an expression?

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  • $\begingroup$ when get we the number $1$ by $8n$? $\endgroup$ – Dr. Sonnhard Graubner Sep 16 '17 at 20:44
  • $\begingroup$ it should be $0$ i believe? $\endgroup$ – K Split X Sep 16 '17 at 20:44
  • $\begingroup$ Perhaps the $1$ is not part of the sequence. For example, perhaps the OP means the sum $$1+\sum_{k=1}^n 8k$$. $\endgroup$ – Franklin Pezzuti Dyer Sep 16 '17 at 20:45
  • $\begingroup$ We define $0$ as a perfect square too, no? $\endgroup$ – K Split X Sep 16 '17 at 20:45
  • $\begingroup$ @Nilknarf that makes a lot of sense actually $\endgroup$ – George Coote Sep 16 '17 at 20:45
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You are saying that the value of the sum $$1+\sum_{k=1}^n 8k$$ is always a perfect square. However, since $$\sum_{k=1}^n k=\frac{n(n+1)}{2}$$ we have $$1+\sum_{k=1}^n 8k=1+8\sum_{k=1}^n k$$ $$1+\sum_{k=1}^n 8k=1+\frac{8n(n+1)}{2}$$ $$1+\sum_{k=1}^n 8k=1+4n(n+1)$$ $$1+\sum_{k=1}^n 8k=1+4n^2+4n$$ $$1+\sum_{k=1}^n 8k=(2n+1)^2$$ and so it always has the value $(2n+1)^2$, which is obviously a perfect square.

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Hint: $$\sum_{i=1}^n i = \frac {n(n+1)} 2$$

And you have the series,

$$1 + \sum_{i=1}^n 8i$$

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you will get $$1+\sum_{k=1}^n8k=1+4n(n+1)$$, can you prove that this is a complete square?

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