A new method of squaring and cubing

Below we demonstrate a simple method to find nth power of a number. Here we’ll take examples to find Square & Cube of a number through Points marked on 2 faces & 3 faces of a Triangular Pyramid respectively.

Firstly, we are finding cube of a number.

1. We take 3 Pyramid faces and mark the Left Side with 1, 2, 3, 4,…….so on.
2. And on the Right Side we are taking the sum of points as shown in Figure-1.

By the Numbers of Points on The Three Faces of a “PYRAMID” in the above Figure-1

We shall prove N³ =N (3N-2) + N (N-1) (N-2).

Now

In Figure-1

No of Points (1, 7, 13, 19, 25……) , It is an A.P Series.

If we do 1³ =1 + (1*0*…..)

2³ = (1+7) + (2*1*0) =8

3³ = (1+7+13) + (3*2*1) =21 + 6 = 27 and so on

N³=(N/2)[2a+(N-1)*d] + C(N,3)

Putting the value a=1 & d=6, we get

N³=(N/2)[2*1+(N-1)*6] + C(N,3)

=(N/2)[2+6N-6] + C(N,3)

=(N/2)[6N-4] + C(N,3)

=N(3N-2) + N(N-1)(N-2)

N³ =N (3N-2) + N (N-1) (N-2) (Hence Proved)

Secondly, we are finding Square of a number.

1. We take 2 Pyramid faces and mark the Left Side with 1, 2, 3, 4,…….so on.
2. And on the Right Side we are taking the sum of points as shown in Figure-2.

By the Numbers of Points on The Two Faces of a “PYRAMID” in the above Figure-2 .

We shall prove N²=N (2N-1) –N (N-1).

Now

In Figure-2

No of Points (1, 5, 9, 13, 17……) , It is an A.P Series.

If we do

1² =1 – (1*0)

2² = (1+5) – (2*1) = 6 – 2 =4

3²= (1+5+9) – (3*2) =15 – 6 = 9

4²= (1+5+9+13) – (4*3) =28 – 12 = 16

N²=(N/2)[2a+(N-1)*d] – C(N,2)

Putting the value a=1 & d=4, we get

N²=(N/2)[2*1+(N-1)*4] – C(N,2)

=(N/2)[2+4N-4] – C(N,2)

=(N/2)[4N-2] – C(N,2)

=N(2N-1) – N(N-1)

Hence, N²=N (2N-1) –N (N-1).