Which summation properties or formulas must be used to evaluate this summation? Check all that
apply

Given:
Summation problem is:
[tex]\sum\limits_{n=1}^{35}n(n^2+1)[/tex]
To find:
The properties or formulas must be used to evaluate the given summation.
Solution:
We have,
[tex]\sum\limits_{n=1}^{35}n(n^2+1)[/tex]
In can be rewritten as:
[tex]\sum\limits_{n=1}^{35}n(n^2+1)=\sum\limits_{n=1}^{35}(n^3+n)[/tex]
[tex]\sum\limits_{n=1}^{35}n(n^2+1)=\sum\limits_{n=1}^{35}n^3+\sum\limits_{n=1}^{35}n[/tex] [tex]\left[\because \sum\limits_{i=1}^n(a_i\pm b_i)=\sum\limits_{i=1}^na_i\pm \sum\limits_{i=1}^nb_i\right][/tex]
[tex]\sum\limits_{n=1}^{35}n(n^2+1)=\left[\dfrac{35(35+1)}{2}\right]^2+\dfrac{35(35+1)}{2}[/tex] [tex]\left[\because \sum\limits_{i=1}^ni=\dfrac{n(n+1)}{2},\sum\limits_{i=1}^ni^3=\left[\dfrac{n(n+1)}{2}\right]^2\right][/tex]
[tex]\sum\limits_{n=1}^{35}n(n^2+1)=\left[\dfrac{35(36)}{2}\right]^2+\dfrac{35(36)}{2}[/tex]
[tex]\sum\limits_{n=1}^{35}n(n^2+1)=\left[630\right]^2+630[/tex]
[tex]\sum\limits_{n=1}^{35}n(n^2+1)=396900+630[/tex]
[tex]\sum\limits_{n=1}^{35}n(n^2+1)=397530[/tex]
Therefore, the properties and formula used are:
[tex]\sum\limits_{i=1}^n(a_i\pm b_i)=\sum\limits_{i=1}^na_i\pm \sum\limits_{i=1}^nb_i,\\\sum\limits_{i=1}^ni=\dfrac{n(n+1)}{2},\\\sum\limits_{i=1}^ni^3=\left[\dfrac{n(n+1)}{2}\right]^2[/tex]
Answer:
Top column right
Both bottom column
Step-by-step explanation: