**How to go back in a table so the common difference is adding each time by 5,7,9 etc**

The **Answer to the Question**

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**Here's the Solution to this Question**

Let us start a sequence from $f(0)=a$. We have next table:

$\begin{matrix} x & 0 & 1 & 2 & \dots & n \\ f(x) & a & a+5 & a+5+7 & \dots & a+\sum\limits_{k=1}^n (2k+3) \end{matrix}$

So, $f(n)-f(n-1)=a+\sum\limits_{k=1}^{n} (2k+3)-a-\sum\limits_{k=1}^{n-1} (2k+3)=2n+3$ and $2n+3 \in \{5,7,9,\dots\}$

Moreover, $f(n)=a+\sum\limits_{k=1}^n (2k+3)=a+2\frac{n(n+1)}{2}+3n=a+n^2+4n$