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I have the following sequence:

$n=1$ $$1$$ $n=2$ $$2+3+4$$ $n=3$ $$5+6+7+8+9$$ $n=4$ $$10+11+12+13+14+15+16$$ $$...$$

So abstracting these sums, we can write:

$n_1+(n_1+1)+...+(n_2-1)+n_2$

, where $n_1$ is the first term an $n_2$ is the last term of the sum.

And the sum of every $n^{th}$ term is: $\frac{1}{2}(n_2-n_1+1)(n_1+n_2)$

Now I want to show that to calculate that the sum up to a certain $n$ value can be calculated as:

$$\frac{1}{2}n^2(n^2+1)$$

But I get stuck in finding this expression from the prior knowledge I described above. I see it is closely related to the formula for the sum of nonnegative integers, but I can't figure out the line of reasoning that brings me to this formula for the sum of the above sequence for an $n^{th}$ value.

I could also try to proof that this formula is correct and I immediately thought of induction. So we can first write $$\frac{1}{2}(n_2-n_1+1)(n_1+n_2)$$ in terms of $n+1$, like: $$\frac{1}{2}(n^2+1-(n+1)^2)(2n+1)$$

Then for the base case, $n=1$, we have $$\frac{1}{2}1^2(1^2+1) = 1$$

Now for the inductive step, I started with:

$$f(n+1) = f(n) + \frac{1}{2}(n^2+1-(n+1)^2)(2n+1)$$

then proceded with the RHS as:

$$\frac{1}{2}n^2(n^2+1)+ \frac{1}{2}(n^2+1-(n+1)^2)(2n+1)$$

Now, to get the inductive step to work, we need to end up with:

$$f(n+1) = \frac{1}{2}(n+1)^2((n+1)^2+1)$$

But I don't see how we can proceed from where I've finished. I can do some algebraic manipulations, but it doesn't brings me to the above equation.

So my questions are:

  1. For the first part: how can I show that the sum of the sequence for a given $n$ is equal to $$\frac{1}{2}n^2(n^2+1)$$

  2. For the second part/the proof by induction part:

  • Is my reasoning and are my calculations up to this point correct (in order to prove the thing that I want to prove)?

  • If yes, how can I continue to get my inductive step to work. And if no, where are my calculations incorrect or when the reasoning is incorrect, how can I prove, perhaps via an alternative route, that the sum up to a certain $n$-value is indeed calculated as:

$$\frac{1}{2}n^2(n^2+1)$$

Steven01123581321
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1 Answers1

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The sum is equal to the sum of integers up to $n^2$ which is trivially given by $\frac{1}{2}n^2(n^2+1)$, then we could prove at first the general result by induction and then use it for this particular case.

Refer also to the related

Edit

Following your way, for the inductive step we have

$$\frac{1}{2}n^2(n^2+1)+\overbrace{(n^2+1)+(n^2+2)+\ldots+(n+1)^2}^{2n+1\: \text{terms}}=$$

$$=\frac{1}{2}n^2(n^2+1)+n^2(2n+1)+\frac12(2n+1)(2n+2)=$$

$$=\frac{1}{2}n^2(n^2+1)+\frac{1}{2}n^2(2n+1)+\frac{1}{2}n^2(2n+1)+\frac12(2n+1)(2n+2)=$$

$$=\frac{1}{2}n^2(n^2+1+2n+1)+\frac{1}{2}(2n+1)(n^2+2n+2)=$$

$$=\frac12 (n^2+2n+1)(n^2+2n+2=\frac12 (n+1)^2+((n+1)^2+1)$$

user
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