118. If BY and XY be given, to find the third term AY, we have only to describe any right-angled triangle on XY as the hypotenuse and make angle APX = angle XPB.

119. Let AX=a, AB = b, and AY= c.

Ihen b = 2ac / a+c ; or, ab + bc = 2ac or,c c= ab / 2a - b = b / 2 - b/a.

When a = b, c = b. When b = 2a, c= ∞. Therefore when X is the middle point of AB, Y is at an infinite distance to the right of B. Y approaches B as X approaches it, and ultimately the three points coincide.

As X moves from the middle of AB to the left, Y moves from an infinite distance on the left towards A, and ultimately X, A, and Y coincide.

120. If E be the middle point of AB,

EX.EY = EA2 = EB2 for all positions of X and Y with reference to A or B.

Each of the two systems of pairs of points X and Y is called a system in involution, the point E being called the center and A or B the focus of the system. The two systems together may be regarded as one system.

121. AX and AY being given, B can be found as follows:

Produce XA and take AC=XA.

Take D the middle point of A Y.

Take CE = DA or AE=DC.

Fig. 43.

Fold through A so that AF may be at right angles to CA Y.

Find F such that DF=DC.

Fold through EF and obtain FB, such that FB is at right angles to EF.

CD is the arithmetic mean between AX and A Y.

AF is the geometric mean between AX and A Y.

AF is also the geometric mean between CD or AE and AB.

Therefore AB is the harmonic mean between AX and A Y

122. The following is a very simple method of finding the harmonic mean between two given lines.

Take AB, CD on the edges of the square equal to the given lines. Fold the diagonals AD, BC and the sides AC, BD of the trapezoid ACDB. Fold through E, the point of intersection of the diagonals, so that FEG may be at right angles to the other sides of the square or parallel to AB and CD. Let FEG cut AC and BD in F and G. Then EG is the harmonic mean between AB and CD.

Fig. 44.

For FE/AB = CE/CB and EG/CD = FE/CD = EB/CB

... FE / AB + EF / CD = CE / CB + EB / CB = 1.

... 1 / AB + 1 / CD = 1 / FE = 2 / FG.

123. The line HK connecting the mid-points of AC and BD is the arithmetic mean between AB and CD.

124. To find the geometric mean, take HL in HK = FG. Fold LM at right angles to HK. Take 0 the mid-point of HK and find M in LM so that OM=OH HM is the geometric mean between AB and CD as well as between FG and HK. The geometric mean between two quantities is thus seen to be the geometric mean between their arithmetic mean and harmonic mean.

Fig. 45.

Summation Of Certain Series

125. To sum the series

1 + 3+5.. .. + (2n - 1).

Divide the given square into a number of equal squares as in Fig. 45. Here we have 49 squares, but the number may be increased as we please.

The number of squares will evidently be a square number, the square of the number of divisions of the sides of the given square.

Let each of the small squares be considered as the unit; the figure formed by A + O + a being called a gnomon.

The numbers of unit squares in each of the gnomons AOa, BOb, etc., are respectively 3, 5, 7, 9, 11, 13.

Therefore the sum of the series 1, 3, 5, 7, 9, 11, 13 is 72.

Generally, 1 + 3 + 5 + .... +(2n - 1) = n2.

Fig. 46.

126. To find the sum of the cubes of the first n natural numbers.

Fold the square into 49 equal squares as in the preceding article, and letter the gnomons. Fill up the squares with numbers as in the multiplication table.

The number in the initial square is 1 = l3.

The sums of the numbers in the gnomons Aa, Bb, etc., are 2 + 4 + 2 = 23, 33, 43, 53, 63, and 73.

The sum of the numbers in the first horizontal row is the sum of the first seven natural numbers. Let us call it s.

Then the sums of the numbers in rows a, b, c, d, etc., are

2s, 3s, 4s, 5s, 6s, and 7s.

Therefore the sum of all the numbers is s(1 + 2 + 3 + 4+5 + 6 + 7)=s2.

Therefore, the sum of the cubes of the first seven natural numbers is equal to the square of the sum of those numbers.

Generally, l3 + 23 + 33 ......+ n3

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

...Σn3 =

For [n•(n + l)]2 - [(n - 1)-n]2

= (n2 + n)2 - (n2 - n)2 = 4n3. Putting n = 1, 2, 3.... in order, we have 4.l3 = (l.2)2 - (0.1)2 4.23 = (2.3)2 - (1.2)2 4.33 = (3.4)2 - (2.3)2

4.n3 =[n•(n+l)]2 - [(n - l)-n]2.

Adding we have

4Σn3 = [n(n + l)]2

... Σn3 =

127. If sn be the sum of the first n natural numbers, sn2 - sn2 - 1 = n3.

128. To sum the series

1.2 + 2.3 + 3.4.... + (n - 1).n.

In Fig. 46, the numbers in the diagonal commencing from 1, are the squares of the natural numbers in order.

The numbers in one gnomon can be subtracted from the corresponding numbers in the succeeding gnomon. By this process we obtain

+ 2[n(n - l) + (n - 2) + (n - 3)....+l]

=n2 + (n - l)2 + 2[l+2.... + (n - l)]

= n2+ (n - l)2 + n(n - 1) = [n - (n - 1)]2 + 3(n - 1)n = 1 +3(n - 1)n. Now ... n3 - (n - 1)3 = 1 + 3(n - 1)n,

... (n - 1)3 - (n - 2)3 = l+3(n - 2) (n -l)

.....................................................

23 - 13 = 1 + 3.2.l

13 - 03 = l + 0.

Hence, by addition, n3 = n+3[l.2 + 2.3+......+ (n - 1).n].

Therefore

1.2+2.3....+(n-1).n = n3 - n /3 = (n-1) n (n+1) / 3

129. To find the sum of the squares of the first n natural numbers.

1.2 + 2.3....+(n - 1).n

= 22 - 2 + 32 - 3........+ n2 - n

= l2+22 + 32.. .. + n2 - (1 +2 + 3.... + n)

= l2 + 22 + 32.... + n2-n(n+1) / 2.

Therefore l2 + 22 + 32....+ n2=(n-1)n(n+1) / 3 +n(n+1) / 2

=n(n+1)

=n(n + l)(2n+l)/ 6.

130. To sum the series l2 + 32 + 52.. .. + (2n - l)2. ... n3 - (n - l)3=n2 +(n - 1)2+n(n - 1), by § 128, = (2n - l)2 - (n - l).n, ... by putting n = 1, 2, 3,.. .. 13 - 03 = 12 - 0.1 23 - 13 = 32 - 1.2 33 - 23 = 52 - 2.3

...........................

...........................

n3 - (n - 1)3 = (2n - l)2 - (n - 1)•n.

Adding, we have n3 = 12 + 32 + 52....... + (2n - 1)2

- [1.2 + 2.3 +3.4... + (n - l).n].

... 12 +32 + 52 ...... + (2n - 1)2

= n3 + n3 - n / 3

= 4n3 - n / 3 = n(2n - 1)(2n+1) / 3