Answer:
the value of the series;
[tex]\sum_{k=1}^{6}(25-k^2) = 59[/tex]
C) 59
Step-by-step explanation:
Recall that;
[tex]\sum_{1}^{n}a_n = a_1+a_2+...+a_n\\[/tex]
Therefore, we can evaluate the series;
[tex]\sum_{k=1}^{6}(25-k^2)[/tex]
by summing the values of the series within that interval.
the values of the series are evaluated by substituting the corresponding values of k into the equation.
[tex]\sum_{k=1}^{6}(25-k^2) =(25-1^2)+(25-2^2)+(25-3^2)+(25-4^2)+(25-5^2)+(25-6^2)\\\sum_{k=1}^{6}(25-k^2) =(25-1)+(25-4)+(25-9)+(25-16)+(25-25)+(25-36)\\\sum_{k=1}^{6}(25-k^2) =24+21+16+9+0+(-11)\\\sum_{k=1}^{6}(25-k^2) = 59\\[/tex]
So, the value of the series;
[tex]\sum_{k=1}^{6}(25-k^2) = 59[/tex]
