The summation of integers with finite integral power (constant for all integers) up to a finite integer can be evaluated. The correct expressions filled would be in the sequence c,b,d,e,a.
How are positive integers summed upto a length with one, two and three degree power of each integer?
Suppose we take the sum of natural numbers (also called positive integers) raised to power 1, 2, and 3, till the nth integer. Then, their sums are given as shown below:
[tex]S_1 = 1 + 2 + \cdots + n = \sum_{i=1}^n i = \dfrac{n(n+1)}{2}\\\\S_2 =1^2 + 2^2 + \cdots + n^2 = \sum_{i=1}^n i^2 = \dfrac{n(n+1)(n+2)}{6}\\\\S_3 = 1^3 + 2^3 + \cdots + n^3 = \sum_{i=1}^n i^3 = \left[\dfrac{n(n+1)}{2}\\\right]^2[/tex]
For this case, evaluating the rest of the two summations:
[tex]\sum_{i=1}^n ca_i = ca_1 + ca_2 + \cdots + ca_n = c(a_1 + a_2 + \cdots + a_n)\\\sum_{i=1}^n ca_i = c\sum_{i=0}^n a_i[/tex]
c was not depending on the variable on which summation is done, which was i. Therefore, it can come out directly.
Thus, whenever we get:
[tex]\sum_x k f(x)[/tex] where k is not depending on the value of x, then we can take it as:
[tex]\sum_x k f(x) = k\left(\sum_x f(x) \right)[/tex] (using the distributive property of multiplication over addition).
Similarly, we get:
[tex]\sum_{i=1}^n c = c + c + \cdots + c \text{ (n times)} = c\times n\\\sum_{i=1}^n ca_i = c.n[/tex]
Thus, the summation of integers with finite integral power (constant for all integers) up to a finite integer can be evaluated. The correct expressions filled would be in the sequence c,b,d,e,a.
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