Respuesta :
Answer:
The sum of the series is Sₙ = n/2 [2·a + (n - 1)·d] where a = 8 and d = 8, therefore 8 + 16 + 24 + ... + 8·n = 4·n·(n + 1)
Step-by-step explanation:
The parameters given are;
8 + 16 + 24 + ... + 8·n = 4·n·(n + 1)
The given series of numbers can be checked to find;
16 - 8 = 24 - 16 = 8
Therefore, the series of numbers is an arithmetic progression with first term = 8, and common difference = 8, we have;
The sum of n terms of an arithmetic progression, Sₙ, is given as follows;
Sₙ = n/2 [2·a + (n - 1)·d]
Where;
a = The first term of the series of numbers = 8
d = The common difference = 8
∴ Sₙ = n/2 × [2×8 + (n - 1)×8] = n [2×8/2 + (n - 1)×8/2] = n × [8 + (n - 1)×4]
Sₙ = n × [8 + (n - 1)×4] = n × [8 + 4·n - 4] = n × [8 - 4 + 4·n] = n × [4 + 4·n]
Sₙ =n × [4 + 4·n] = 4 × n×(n + 1) = 4·n·(n + 1).
