Respuesta :
Answer:
[tex]\sum\limits_{i=1}^{n} c \cdot a_i \ is \ equal \ to \ c = \ c \cdot \sum\limits_{i=1}^{n} a_i[/tex]
[tex]\sum\limits_{i=1}^{n} i \ is \ equal \ to \ b = \dfrac{n \cdot (n + 1)}{2}[/tex]
[tex]\sum\limits_{i=1}^{n} c \ is \ equal \ to \ d = c \cdot n[/tex]
[tex]\sum\limits_{i=1}^{n} i^3 \ is \ equal \ to \ e = \left [ \dfrac{n \cdot (n + 1)}{2} \right]^2[/tex]
[tex]\sum\limits_{i=1}^{n} i^2 \ is \ equal \ to \ a = \dfrac{n \cdot (n + 1) \cdot (2 \cdot n + 1)}{6}[/tex]
Step-by-step explanation:
1) The sum of a series of the cubes of numbers 'i' is given as follows;
Sₙ = n/2 × (1st term + Last term)
∴ Sₙ = n/2 × (1 + n) = n·(n + 1)/2
It can be shown that the sum of a series of the cubes of numbers 'i²' is given as follows;
Sₙ = (n·(n + 1)·(2·n + 1))/6
It can be also be shown that the sum of a series of the cubes of numbers 'i³' is given as follows;
Sₙ = (n·(n + 1)/2)²
Therefore, we have;
[tex]\sum\limits_{i=1}^{n} c \cdot a_i= c \cdot \sum\limits_{i=1}^{n} a_i = c[/tex]
[tex]\sum\limits_{i=1}^{n} i = \dfrac{n \cdot (n + 1)}{2} = b[/tex]
[tex]\sum\limits_{i=1}^{n} c = c \cdot n = d[/tex]
[tex]\sum\limits_{i=1}^{n} i^3 = \left [ \dfrac{n \cdot (n + 1)}{2} \right]^2 = e[/tex]
[tex]\sum\limits_{i=1}^{n} i^2 = \dfrac{n \cdot (n + 1) \cdot (2 \cdot n + 1)}{6} = a[/tex].
