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Scilla [17]
3 years ago
12

Evaluate the series

Mathematics
1 answer:
dalvyx [7]3 years ago
6 0

Answer:

the value of the series;

\sum_{k=1}^{6}(25-k^2) = 59

C) 59

Step-by-step explanation:

Recall that;

\sum_{1}^{n}a_n = a_1+a_2+...+a_n\\

Therefore, we can evaluate the series;

\sum_{k=1}^{6}(25-k^2)

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.

\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\\

So, the value of the series;

\sum_{k=1}^{6}(25-k^2) = 59

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