60 Chapter 6

6.5 Further identities

Let us now consider several further questions concerning the identity of the previous prob-

lem.

5

2

− 4

2

+ 3

3

− 2

2

+ 1

2

1

2

3

4

5

6



6

2



1 + 2 + 3 + 4 + 5

5 ·6/2

In the exercises, you will be asked to explain how these diagrams establish the identities:

n



k=1

(−1)

n−k

k

2

=



n + 1

2



=

n



k=1

k =

n(n + 1)

2

.

6.6 Sum of odd numbers

Next, consider this diagram as a proof that the sum of the first n odd numbers is n

2

:

1

3

5

7

9

11

13

15

1 + 3 + 5 + ···+ (2n − 1) = n

2

Can you see how the diagram proves the identity? The large square, with area n

2

,is

partitioned into bent sections of size 1, size 3, size 5, and so on, for the first n odd numbers.

So the sum of those areas is equal to n

2

, the total area. Do you prefer this proof to the

induction argument we gave in theorem 23?