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- Marked by Teachers essays 2
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- Level: GCSE
- Questions: 75
is (n-19) as it has been decreased by 1. The cell to the right of (n-18) is (n-17) as it is 1 more than (n-18). When these 5 terms are added together I get: (n) + (n-9) + (n-17) + (n-18) + (n-19) = 5n - 63 The calculation above shows that the sum of the 5 terms within the T-shape is 5n - 63, therefore I can make a proper formula: T = 5n - 63 where T is the T-total and n is the T-number T-number (n) T-total (T) T-total using formula (5n-63) 20 37 (5x20)= 100 100-63 = 37 26 67 (5x26)= 130 130-63 = 67 50 187 (5x50)= 250 250-63 = 187 80 337
- Word count: 2995
* When the T-Number is even, the T-Total is even. I will now find a rule which links the T-number with the T-Total: n+(n-8)+(n-16)+ (n-18)+(n-17) =5n-56 When n=36 =(5x36)-56=124 Testing: 19+20+21+28+36=124 As you can see my rule has worked. T-Totals - Any sized Grid I will now find the general rule for any sized grid, which links the T-Number with the T-Total. n+(n-G)+(n-2G) +(n-2G-1)+(n-2G+1) = 5n-7G When n=65, and G=10 =(5x65)-(7x10)= 255 44+45+46+55+65= 225 As you can see my rule has worked. Translation: If I translate the T 3 Vectors right, it will become: 22+23+24+33+43= 145 25+26+27+36+46= 160 T-Number=43 T-Number=46 T-Total= 145 T-Total=160 * The T-Total has increased by 15.
- Word count: 1156