19. \( 1+2+3+4+5+\ldots+n=\frac{n(n+1)}{2} \) (a) i) Show that this formula is true for the sum of the first 8 natural numbers. ii) Find the sum of the first 400 natural numbers. (b) i) Show that \( 2+4+6+8+\cdots+2 n=n(n+1) \). ii) Find the sum of the first 200 even numbers. iii) Find the sum of the first 200 odd numbers. (c) i) Use the formula at the beginning of the question to find the sum of the first \( 2 n \) natural numbers. ii) Find a formula, in its simplest form, for \( 1+3+5+7+9+\cdots+(2 n-1) \).

Let's solve each part of the problem step by step. ### Part (a) #### (i) Show that the formula is true for the sum of the first 8 natural numbers. We need to calculate \( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 \). Using the formula: \[ \frac{n(n+1)}{2} \quad \text{where } n = 8 \] \[ \frac{8(8+1)}{2} = \frac{8 \times 9}{2} = \frac{72}{2} = 36 \] Now, let's calculate the sum directly: \[ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 \] Thus, the formula is verified for \( n = 8 \). #### (ii) Find the sum of the first 400 natural numbers. Using the formula: \[ \frac{n(n+1)}{2} \quad \text{where } n = 400 \] \[ \frac{400(400+1)}{2} = \frac{400 \times 401}{2} = \frac{160400}{2} = 80200 \] So, the sum of the first 400 natural numbers is \( 80200 \). ### Part (b) #### (i) Show that \( 2 + 4 + 6 + 8 + \cdots + 2n = n(n+1) \). The sum can be factored as: \[ 2(1 + 2 + 3 + \ldots + n) = 2 \cdot \frac{n(n+1)}{2} = n(n+1) \] Thus, the formula is verified. #### (ii) Find the sum of the first 200 even numbers. Using the formula: \[ n(n+1) \quad \text{where } n = 200 \] \[ 200(200+1) = 200 \times 201 = 40200 \] So, the sum of the first 200 even numbers is \( 40200 \). #### (iii) Find the sum of the first 200 odd numbers. The sum of the first \( n \) odd numbers is given by the formula \( n^2 \). For \( n = 200 \): \[ 200^2 = 40000 \] So, the sum of the first 200 odd numbers is \( 40000 \). ### Part (c) #### (i) Use the formula at the beginning of the question to find the sum of the first \( 2n \) natural numbers. Using the formula: \[ \frac{n(n+1)}{2} \quad \text{where } n = 2n \] \[ \frac{2n(2n+1)}{2} = n(2n+1) \] So, the sum of the first \( 2n \) natural numbers is \( n(2n+1) \). #### (ii) Find a formula, in its simplest form, for \( 1 + 3 + 5 + 7 + 9 + \cdots + (2n-1) \). The sum of the first \( n \) odd numbers is given by: \[ n^2 \] Thus, the formula for \( 1 + 3 + 5 + 7 + 9 + \cdots + (2n-1) \) is \( n^2 \). ### Summary of Results - The sum of the first 8 natural numbers is \( 36 \). - The sum of the first 400 natural numbers is \( 80200 \). - The sum of the first 200 even numbers is \( 40200 \). - The sum of the first 200 odd numbers is \( 40000 \). - The sum of the first \( 2n \) natural numbers is \( n(2n+1) \). - The sum of the first \( n \) odd numbers is \( n^2 \).