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Introduction

Candidate Name – Mustafa Ali Halawa                                 Candidate Number -

1+4=1*1+2*2

I will construct 2D cross shape but explain them in relation to 3D cross shapes.

The Third 3D Cross shape has 13 cubes in its middle layer which is based on its corresponding 2D cross shape.

Therefore:   13=4+9

=2*2+3*3

The fourth 3D cross shape has 25 cubes in its middle layer which is based on its corresponding 2D cross shape.       25=16+9

=4*4+3*3

Middle

41=16+25

=4*4+5*5 The sixth 3D cross shape has 61 cubes in its middle layer which is based on its corresponding 2D cross shape.                 61=25+36 =5*5+6*6

By tabulating all the above result I can derive the general result.

 Number/level (N) Mn(layer of cubes in the middle layer) 1 1=1=1*1=0*0+1*1 2 5=1+4=1*1+2*2 3 13=4+9=2*2+3*3 4 25=9+16=3*3+4*4 5 41=16+25=4*4+5*5 6 61=25+36=5*5+6*6

The first column values of the table were always 1 less than the corresponding cross shape where as the second column values are the cross shape numbers broken up.

Conclusion

My GCSE curriculum   does not contain any method of finding this sum of squares which is why I have looked elsewhere to prove this result. I have taken from the book “Further mathematics 1” by Geoff Mannall and Michael Kenwood.

12+22+32+…...+n2=nΣr2

r=1

Where Σ (sigma) is the summation of terms using the identity 24r2+2Ξ (2r+1)3-(2r-1)3 and take r=1, 2, 3,…..n

When r=1; 24(1)2+2=33-13

r=2; 24(2)2+2=53-33

r=3; 24(3)2+2=73-53

r=n-1; 24(n-1)2+2= (2n-1)3-(2n-3)3

r=n; 24n2+2= (2n+1)3-(2n-3)3

By adding all the above right terms, the result is (2n+1)3-13 which is equal to

8n3+12n2+6n

Centre Name – St. Mary’s Catholic High School                Centre Number – 91530

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