Maths Presentation Essay

353 WordsNov 9, 20112 Pages
If every pupil call each other on a mobile phone, how many calls will there be altogether when there are 910 pupils? Ok. So if two people rang each other then it would look like this: =pupil 2 pupils=1 contacts =contact This shows that two people can only ring each other once. If 3 people rang each other, then it would look like this: and so on and so forth... 3 pupils= 3 contacts 4 pupils= 6 contacts If I put all the combinations in up to six in a sequence then it would look like this: 2 3 4 5 6>>>>>>>>>> nth term 1 3 6 10 15>>>>>>>> sequence 2 3 4 5 >>>>>>>>>>>difference 1 1 1 To find the nth term “an2+bn+c” II. 4a+2b+c=1 III. 9a+3b+c=3 IV. 16a+4b+c=6 Now I am going to cancel out the c by taking away equation II from III: V. (9a+3b+c=6) – (4a+2b+c=1) = (5a+b=2) Then I will do the same for equation III and IV: VI. (16a+4b+c=6) – (9a+3b+c=3) = 7a+b=3 Now to work out what “a” is, I will minus equation V from equation VI. VII. (7a+b=3) – (5a+b=2) = 2a=1 a=0.5 upon finding out that “a” equals 0.5, I can work out that... (5x0.5)+b=2 2.5+b=2 ...b=-0.5 So now I will go back to equation II to work out what “c” is: 4x0.5+2x-0.5+c=1 2+ (-1)+c=1 C=0 So the formula is: 0.5n2+ (-0.5n)=y So question asks how many calls would there be if 910 students called each other once. Now I can work out what it is by replacing n for 910: 0.5n2+ (-0.5n)=n(n+1)2=910(910+1)2=414

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