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I have the following equation which is what f is equals to, and am told to write the recursive definition for it. I know that the base case is simply: when n = 0, we can say that the function f = 0^2+2(0)+1 = 1

However, for the recursive definition I am very stuck, as I do not know how to define it. I believe the mathematical formula for this is $\frac{n(n+1)(2n+1)}{6}$, yet I am unclear how to translate this. $$\sum_{i=1}^n i^2+2i+1$$

FrenchToast
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  • https://math.stackexchange.com/questions/3850288 – 311411 Feb 07 '22 at 01:27
  • You already have the base case, now you just need to write a formula for $f(n)$ in terms of $f(n-1)$ and $n$. What happens when you stop the sum early? – AHusain Feb 07 '22 at 01:27
  • "I believe the mathematical formula for this is..." $;-;$ That doesn't work even for $,n=0,$ or $,n=1,$. What you have there is $,\sum_{i=1}^n\left(i+1\right)^2=\sum_{i=2}^{n+1}i^2,$. – dxiv Feb 07 '22 at 01:41
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    There's no such thing as the recursive definition for $f$; there are infinitely many. Here's one: $f(n)=f(n-1)+n^2+2n+1$. – Gerry Myerson Feb 07 '22 at 01:47
  • For what it's worth, I suspect that the problem composer intends a response consistent with the comment of @GerryMyerson. – user2661923 Feb 07 '22 at 03:39
  • @GerryMyerson Thanks greatly, I had arrived at this solution earlier but thought it was too simple to be correct. Everyones answers are appreciated. – FrenchToast Feb 07 '22 at 05:19

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