Question

1. Find the sum of the following AP (i) 2, 4, 6, ... to n terms (ii) 7, 71, 72, 73, ... to 50 terms (iii) a, (a + b), (a + 2b), ... to n terms (iv) x + y, x-y, x-3y,... to 20 terms (v) (a - b)², (a² + b2), (a + b)², ... to n terms.​

Answer 1

Let $S_n$ be the sum of the first $n$ natural numbers.

$S_n = 1 + 2 + 3 + \cdots + n$

We can reverse the order of terms to write

$S_n = n + (n-1) + (n-2) + \cdots + 1$

so that doubling up, we get

$2S_n = (1 + n) + (2 + (n-1)) + (3 + (n-2)) + \cdots + (n + 1)$

$2S_n = (n + 1) + (n + 1) + (n + 1) + \cdots + (n + 1)$

and since there are $n$ terms in $S_n$, it follows that

$2S_n = n(n+1) \implies S_n = \dfrac{n(n+1)}2$

Now, for an arithmetic sequence starting with $a_0$ and having common difference $d$ between consecutive terms, the sum of the first $n$ terms of this sequence is

$a_0 + (a_0 + d) + (a_0 + 2d) + \cdots + (a_0 + (n-1)d) \\\\ ~~~~~~~~ = (a_0 - d + d) + (a_0 - d + 2d) + (a_0 - d + 3d) + \cdots + (a_0 - d + nd) \\\\ ~~~~~~~~ = (a_0 - d) n + d S_n \\\\ ~~~~~~~~ = \dfrac d2 n^2 + \left(a_0 - \dfrac d2\right) n$

(i) This sum is simply

$2 + 4 + 6 + \cdots + 2n \\\\ ~~~~~~~~ = 2 (1 + 2 + 3 + \cdots + n) \\\\ ~~~~~~~~ = 2 S_n \\\\ ~~~~~~~~ = \boxed{n(n+1)}$

(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

$70 + 71 + 72 + \cdots + 119 \\\\ ~~~~~~~~ = \dfrac12\cdot50^2 + \left(70 - \dfrac12\right)\cdot50 \\\\ ~~~~~~~~ = \boxed{4725}$

(iii) The first term is $a$ and the common difference is $b$, so

$a + (a + b) + (a + 2b) + \cdots + (a + b(n-1)) \\\\ ~~~~~~~~ = \boxed{\dfrac b2 n^2 + \left(a - \dfrac b2\right)n}$

(iv) The first term is $x+y$ and the common difference is $-2y$. Then the 20th term in the sequence is $x+y - 19\cdot2y= x-37y$, so

$(x + y) + (x - y) + (x - 3y) + \cdots + (x - 37y) \\\\ ~~~~~~~~ = -\dfrac{2y}2\cdot20^2 + \left(x + y + \dfrac{2y}2\right)\cdot20 \\\\ ~~~~~~~~ = \boxed{20x - 360y}$

(v) Note that the first term is

$(a - b)^2 = a^2 + b^2 - 2ab$

so that the common difference is $2ab$. Then

$(a - b)^2 + (a^2 + b^2) + (a + b)^2 + \cdots + (a^2 + b^2 - 2ab + 2ab(n-1)) \\\\ ~~~~~~~~ = \dfrac{2ab}2 n^2 + \left(a^2 + b^2 - 2ab - \dfrac{2ab}2\right) n \\\\ ~~~~~~~~ = \boxed{abn^2 + (a^2 - 3ab + b^2)n}$