Respuesta :
Let [tex]S_n[/tex] be the sum of the first [tex]n[/tex] natural numbers.
[tex]S_n = 1 + 2 + 3 + \cdots + n[/tex]
We can reverse the order of terms to write
[tex]S_n = n + (n-1) + (n-2) + \cdots + 1[/tex]
so that doubling up, we get
[tex]2S_n = (1 + n) + (2 + (n-1)) + (3 + (n-2)) + \cdots + (n + 1)[/tex]
[tex]2S_n = (n + 1) + (n + 1) + (n + 1) + \cdots + (n + 1)[/tex]
and since there are [tex]n[/tex] terms in [tex]S_n[/tex], it follows that
[tex]2S_n = n(n+1) \implies S_n = \dfrac{n(n+1)}2[/tex]
Now, for an arithmetic sequence starting with [tex]a_0[/tex] and having common difference [tex]d[/tex] between consecutive terms, the sum of the first [tex]n[/tex] terms of this sequence is
[tex]a_0 + (a_0 + d) + (a_0 + 2d) + \cdots + (a_0 + (n-1)d) \\\\ ~~~~~~~~ = (a_0 - d + d) + (a_0 - d + 2d) + (a_0 - d + 3d) + \cdots + (a_0 - d + nd) \\\\ ~~~~~~~~ = (a_0 - d) n + d S_n \\\\ ~~~~~~~~ = \dfrac d2 n^2 + \left(a_0 - \dfrac d2\right) n[/tex]
(i) This sum is simply
[tex]2 + 4 + 6 + \cdots + 2n \\\\ ~~~~~~~~ = 2 (1 + 2 + 3 + \cdots + n) \\\\ ~~~~~~~~ = 2 S_n \\\\ ~~~~~~~~ = \boxed{n(n+1)}[/tex]
(ii) The first term in this sequence should be 70 if it's arithmetic. The common difference is then 1, the 50th term is 70 + (50 - 1)•1 = 119, so
[tex]70 + 71 + 72 + \cdots + 119 \\\\ ~~~~~~~~ = \dfrac12\cdot50^2 + \left(70 - \dfrac12\right)\cdot50 \\\\ ~~~~~~~~ = \boxed{4725}[/tex]
(iii) The first term is [tex]a[/tex] and the common difference is [tex]b[/tex], so
[tex]a + (a + b) + (a + 2b) + \cdots + (a + b(n-1)) \\\\ ~~~~~~~~ = \boxed{\dfrac b2 n^2 + \left(a - \dfrac b2\right)n}[/tex]
(iv) The first term is [tex]x+y[/tex] and the common difference is [tex]-2y[/tex]. Then the 20th term in the sequence is [tex]x+y - 19\cdot2y= x-37y[/tex], so
[tex](x + y) + (x - y) + (x - 3y) + \cdots + (x - 37y) \\\\ ~~~~~~~~ = -\dfrac{2y}2\cdot20^2 + \left(x + y + \dfrac{2y}2\right)\cdot20 \\\\ ~~~~~~~~ = \boxed{20x - 360y}[/tex]
(v) Note that the first term is
[tex](a - b)^2 = a^2 + b^2 - 2ab[/tex]
so that the common difference is [tex]2ab[/tex]. Then
[tex](a - b)^2 + (a^2 + b^2) + (a + b)^2 + \cdots + (a^2 + b^2 - 2ab + 2ab(n-1)) \\\\ ~~~~~~~~ = \dfrac{2ab}2 n^2 + \left(a^2 + b^2 - 2ab - \dfrac{2ab}2\right) n \\\\ ~~~~~~~~ = \boxed{abn^2 + (a^2 - 3ab + b^2)n}[/tex]