Proofs Without Words - 2

Each of the coloured regions have areas equal to consecutive odd numbers beginning with 1,

while the sum of their areas is the number of distinct areas squared,

thus from an algebraic perspective,

1+3+5+7+9+11+...+(2n-1)

=(1+2+3+4+5+...+n)

+(0+1+2+3+4+...+(n-1))

= (1/2)(n)(n+1) + (1/2)(n-1)(n)

= (n^2)/2 + n/2 + (n^2)/2 - n/2

= n^2

1+3+5+.....+2n-1 = n^2

BI 2n-1 = n^2 2-2 = 1 1 = 1

HI =

For all numbers n>= 1 is equal to 1+3+5.... +2n-1 = n^2

PI = 1+3+5+...+2n-1+2(n+1)-1 = n^2 + (2n+1) - 1 = n^2 + 2n+2 -1
= n^2+2n + 1 = (n+1)^2

This is an AP with common difference 2, Sn = n/2[1st term +last term] = n/2(1 + 2n-1) = n/2x2n= n^2