Question

When the first n positive integers are added together, their sum is given by
1
2n(n + 1).
(i) Demonstrate that this result holds for the case n = 5.
(ii) Find the value of n for which the sum is 105.
(iii) What is the smallest value of n for which the sum exceeds 1000?

Answer 1

Step-by-step explanation:

if n=5 then

2(5)(5+1)

10(6)

60

if sum is 105 then

2n(n+1)=105

n(n+1)=105/2

n^2+n=52.5

n+n=√52.5

2n=7.24

n=7.24/2

n=3.65

smallest value of n

2n(n+1)=1000

n(n+1)=1000/2

n^2+n=500

n+n=√500

2n=22.36

n=22.36

n=11.18