Let $x_i$ represent the $i^{th}$ term in the quadratic sequence and $d_i$ is the first difference between neighbouring terms in the sequence, where $i \in \mathbb{N}$

Basically to get the next term, you need to add $d_i$ to the previous term, which can cen written mathematically as:

$$x_{i+1} = x_i+d_i$$

Given that $d_1=1 ; d_2=3 ;d_3=5$ we acn calculate the 2nd difference of the sequence.

We expect the second difference of the sequence to have a common difference (all differences are equal), by definition of "Quadratic Sequence".

$3-1=2$

$5-3=2$

Thus our sequence is really quadratic because we have a common difference of 2 in the 2nd differences.

The first differences, $1,3,5$ form an arithmetic sequence and their first term, $d_1=1$, so:

$$d_i=1+2(i - 1)$$

$$=1 + 2i-2$$

therefore

$$d_i=2i-1$$

To get $d_4,d_5,d_6$ you can consecutively add the 2nd difference of to to get $d_4=7 ; d_5=9; d_6=11$ dont forget that these are the first differences and we will use then to get the terms of the original quadratic sequence.

To get $x_5$ and $x_6$, we can use the fact that $x_7=35$ to calculate them

$x_7=x_6+11$ i.e. $35 =x_6 +11$ which means that $x_6=24$

we also know that $x_5+9=x_6$ i.e. $x_5+9=24$, therefore $x_5=15$

Now to get a general formula for the quadratic sequence, we can calculate $x_1$. Remember that the first difference form an arithmetic sequence, so we know how to calculate the first $k$ first differences that have to be added to get $k+1$ from $x_1$

So

$$d_1+d_2+...+d_k$$

$$= \dfrac{k(d_1+d_k)}{2}$$

$$=\dfrac{k(1+2k-1)}{2}$$

$$=\dfrac{k(2k)}{2}$$

$$=k^2$$

therefore

$$x_{k+1} = x_{1} +d_1+d_2+...+d_k=x_1+k^2$$

so,

$x_{6+1}=x_1+6^2$ but we know that $x_7 =35$ hence $35=x_1+36$ which means $x_1=-1$

To get the general term, $x_i$, we can use the formula $x_{k+1}=x_1+k^2 when $k=i-1$

so

$$x_i=x_1+(i-1)^2$$

substituting,

$$x_i= -1+(i^2-2i+1)$$

so the equation of the quadratic sequence is:

$$x_i=i^2-2i$$