1+10 =11
n+n+n = 3n
End formula =3n+11
After finding my formula algebraically as well I can see that the formula should be correct. However just to make sure I will test the formula using 3 different numbers.
2 step stair : Testing
By using my formula 3n+11 I’m going to test this formula in three different positions referring to the number in the left hand corner.
I am firstly going to test the position 27 on a two step stair in a
10 – by – 10 grid:
Formula = 3n+11
= 3 x 2 +11
= 92
Now I will see if the total number step adds up to 92. This will only work if the formula is accurate.
27+28+37 = 92
I have tested position 27 in a 10 –by – 10 grid, on a 2 step stair. After testing this I can therefore say that my formula has worked. So this shows me that my calculations which I have carried out are accurate.
Now I am going to test the position 75 on a two step stair in a
10 – by – 10 grid.
Formula = 3n+11
= 3 x 75 +11
= 236
Now I will see if the total number step adds up to 92. This will only work if the formula is accurate.
75+76+85 = 236
I have tested position 75 in a 10 –by – 10 grid, on a 2 step stair. After testing this I can therefore say that my formula has worked. So this shows me that my calculations which I have carried out are accurate.
Now I am going to test the position 47 on a two step stair in a
10 – by – 10 grid.
Formula = 3n+11
= 3 x 47 +11
= 152
Now I will see if the total number step adds up to 152. This will only work if the formula is accurate.
47+48+57 = 152
I have tested position 47 in a 10 –by – 10 grid, on a 2 step stair. After testing this I can therefore say that my formula has worked. So this shows me that my calculations which I have carried out are accurate.
I have tested out three different numbers within the 10 – by – 10 number grid on a 2 step stair and therefore I can now say that my formula has worked out throughout working out the calculations for the 2 step stair. For that reason I would say my formula is accurate.
3-step stair:
I will now go about investigating the 3-step stair on a 10 –by – 10 grids. I will firstly investigate this numerically. I’m going to start with the number one and I will move one space to the right. I am then going to find the nth term formula.
Position One :
Total of position 1:
1+2+3+11+12+21 = 50
Position Two :
Total of position 2:
2+3+4+12+13+22 = 56
Position Three :
Total of position 3:
3+4+5+13+14+23 =62
Position Four :
Total of position 4:
4+5+6+14+15+24 = 68
Position Five :
Total of position 5:
5+6+7+15+16+25 = 74
Position Six :
Total of position 6:
6+7+8+16+17+26 = 80
Position Seven :
Total of position 7:
7+8+9+17+18+27 = 86
Now that I have found out the totals of the first seven steps, I will be able to find the formula numerically for the 3-step stair:
nth term:
Rule = +6
nth Term = 6n+b
1st term 50= 6(1) +b
50= 6+b
b= 50-6
b= 44
nth term = 6n+44
Now I have worked out the formula numerically and also by using the value of position 1 and also by finding the difference. I have both times got the same formula 6n+44. Now I will see if this gives me the value of position 3.
Formula = 6n+44
Position 5 = 6 x 3+44
This is correct as in the total for position 3, which I worked out previously and I got the answer 26. This shows me that I have got the right formula as it works.
I’m now going to find out the formula algebraically and see if I get the same formula. If they both mach then it is accurate.
n+n+n+n+n+n =6
1+2+10+11+20 = 44
End formula = 6n+44
After finding my formula algebraically as well I can see that the formula should be accurate. However just to make sure I will test the formula using 3 different numbers.
3-step stair: Testing
By using my formula 6n+44 I’m going to test this formula in three different positions referring to the number in the left hand corner.
I am firstly going to test the position 48 on a two step stair in a
10 –by – 10 grids:
Formula = 6n+44
= 3 x 48+11
= 332
Now I will see if the total number step adds up to 332. This will only work if the formula is correct.
48+49+50+58+59+68 = 332
I have tested position 48 in a 10 –by – 10 grid, on a 3 step stair. After testing this I can therefore say that my formula has worked. So this shows me that my calculations, which I have carried out, are accurate.
Now going to test the position 22 on a two step stair in a
10 –by – 10 grids:
Formula = 6n+44
= 6 x 22 +44
= 176
Now I will see if the total number step adds up to 176. This will only work if the formula is correct.
22+23+32+33+42 = 176
I have tested position 22 in a 10 –by – 10 grids, on a 3-step stair. After testing this I can therefore say that my formula has worked. So this shows me that my calculations that I have carried out are accurate.
Now I am going to test the position 58 on a two step stair in a
10 –by – 10 grids:
Formula = 6n+44
= 6x 58 +44
= 392
Now I will see if the total number step adds up to 392. This will only work if the formula is accurate.
58+59+60+68+69+78 = 392
I have tested position 58 in a 10 –by – 10 grids, on a 3-step stair. After testing this I can therefore say that my formula has worked. So this shows me that my calculations that I have carried out are accurate.
I have tested out three different numbers within the 10 – by – 10 number grid on a 3 step stair and therefore I can now say that my formula has worked out throughout working out the calculations for the 3 step stair. For that reason I would say my formula is accurate.
4-step stair:
I will now go about investigating the 4-step stair on a 10 –by – 10 grids. I will firstly investigate this numerically. I’m going to start with the number one and I will move one space to the right. I am then going to find the nth term formula.
Position One :
Total of position 1:
1+2+3+4+11+12+13+21+22+31 = 120
Position Two :
Total of position 2:
2+3+4+5+12+13+14+22+23+32 = 130
Position Three :
Total of position 3:
3+4+5+6+13+14+15+23+24+33 = 140
Position Four :
Total of position 4:
4+5+6+7+14+15+16+24+25+34 = 150
Position Five :
Total of position 5:
5+6+7+8+15+16+17+25+26+35 = 160
Position Six :
Total of position 6:
6+7+8+9+16+17+18+26+27+36 = 170
Position Seven :
Total of position seven:
7+8+9+10+17+18+19+27+28+37 = 180
Now that I have found out the totals of the first seven steps, I will be able to find the formula numerically for the 4-step stair:
nth term:
Rule = +10
nth Term = 10n+b
1st term 120= 10(1) +b
120= 10+b
b= 120-10
b= 110
nth term = 10n+110
Now I have worked out the formula numerically and also by using the value of position 1 and also by finding the difference. I have both times
Got the same formula 10n+110. Now I will see if this gives me the value of position 6.
Formula = 10n+110
Position 5 = 10 x 6+110 = 170
This is correct as in the total for position 6, which I worked out previously and I got the answer 170. This shows me that I have got the correct formula as it works.
I’m now going to find out the formula algebraically and see if I get the same formula. If they both mach then it is accurate.
n+n+n+n+n+n =10
1+2+10+11+20 = 110
End formula = 10n+110
After finding my formula algebraically as well I can see that the formula should be accurate. However just to make sure I will test the formula using 3 different numbers.
4-step stair: Testing
By using my formula 10n+110 I’m going to test this formula in three different positions referring to the number in the left hand corner.
I am firstly going to test the position 44 on a two step stair in a
10 –by – 10 grids:
Formula = 10n+110
= 3 x 44+110
= 550
Now I will see if the total number step adds up to 550. This will only work if the formula is accurate.
44+45+46+47+54+55+56+64+65+74 = 550
I have tested position 44 in a 10 –by – 10 grid, on a 4 step stair. After testing this I can therefore say that my formula has worked. So this shows me that my calculations that I have carried out are accurate.
Now going to test the position 67 on a four step stair in a
10 –by – 10 grids:
Formula = 10n+110
= 10 x 67 +110
= 780
Now I will see if the total number step adds up to 780. This will only work if the formula is accurate.
67+68+69+70+77+78+79+87+88+97 = 780
I have tested position 31 in a 10 –by – 10 grid, on a 4 step stair. After testing this I can therefore say that my formula has worked. So this shows me that my calculations, which I have carried out, are accurate.
Now I am going to test the position 31 on a four step stair in a
10 –by – 10 grids:
Formula = 10n+110
=10 x 31 +110
= 420
Now I will see if the total number step adds up to 420. This will only work if the formula is accurate.
31+32+33+34+41+42+43+51+52+61 = 420
I have tested position 31 in a 10 –by – 10 grid, on a 4 step stair. After testing this I can therefore say that my formula has worked. So this shows me that my calculations that I have carried out are accurate.
I have tested out three different numbers within the 10 – by – 10 number grid on a 3 step stair and therefore I can now say that my formula has worked out throughout working out the calculations for the 3 step stair. For that reason I would say my formula is accurate.
For this second part of my coursework I will be changing the sizes of the number grids and the shapes. Firstly I am going to use the same
10 – by –10 grid and change the shapes.
Firstly I am going to use a C shape to investigate this formula.
Position One :
Total of position 1:
1+2+3+11+21+22+23 = 83
Position Two :
Total of position 2:
2+3+4+11+21+22+23 = 86
Position Three :
Total of position 3:
3+4+5+13+23+24+25 = 97
Position Four :
Total of position 4:
4+5+6+14+24+25+26 =
Position Five :
Total of position 5:
5+6+7+15+25+26+27 = 111
Position Six :
Total of position 6:
6+7+8+16+26+27+28 = 118
Position Seven :
Total of position 7:
7+8+9+17+27+28+29 = 125
Now that I have found out the totals of the first seven steps, I will be able to find the formula numerically for the 4-step stair:
nth term:
Rule = +7
nth Term = 7n+b
1st term 83= 7(1) +b
83= 7+b
b= 83-7
b= 76
nth term = 7+76
Now I have worked out the formula numerically and also by using the value of position 1 and also by finding the difference. I have both times
Got the same formula 10n+110. Now I will see if this gives me the value of position 6.
Formula = 10n+110
Position 5 = 10 x 6+110 = 170
