1 + 2 + 3 + 4 + 5 + .................................. + n = n(n +1)
10
2
(a) (i) Show that this formula is true for the sum of the first 8 natural numbers. [2]
(ii) Find the sum of the first 400 natural numbers. [1]
(b) (i) Show that 2 + 4 + 6 + 8 + ..................... + 2n = n(n + 1). [1]
(ii) Find the sum of the first 200 even numbers. [1]
(iii) Find the sum of the first 200 odd numbers. [1]
(c) (i) Use the formula at the beginning of the question to find the sum of the first 2n natural
numbers. [1]
(ii) Find a formula, in its simplest form, for
1 + 3 + 5 + 7 + 9 + ........................ + (2n – 1).
[2]
Show your working.
ai)1+2+3+4+5+6+7+8=6=8/2(8+1)
ii)400(400+1)/2=80200
bi)AMS=2* the sum of the first 200 odd numbers=2*200(201)/2=40200
ii)ANS = sum of first 400 numbers- sum of first200 even numbers=80200-40200=40000
ci)2n(2n+1)/2=2n^2+n
ii)ANS=sum of firast wn numbers - sum of first 2n even numbers=2n^2+n- 2* sum of first n numbers=2n^2+n-2n(n+1)/2=n^2
