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# My Mathematics Passion – Piyush Goel

Piyush, born on 10th Feb, 1967, Aquarian belongs to a middle class family,   elder son of Dr. Devender Kumar Goel and mother  Ravikanta. He is Diploma Mechanical Engineering , Diploma in Material Management, Diploma in Vastu Shastra , creative, believe in God too much, believe in Love & Friendship.

Piyush Goel has a hobby of collection he has Stamps Collection, Autographs Collection, Collection,Pen Collection ,First Day Cover Collections etc.
Piyush Goel also completed  World First Five Mirror Imaged Books in Five Ways, Shreemadbhagvadgita with Pen, Madhushala with Needle, Gitanjali with Mehndi Cone, Piyush Vani with Iron Nail and Fabric Cone Liner and Panchtantra with Carbon Paper and Meri Ekyavan Kavitayen of Atal Ji with Wooden Pen.
His Work in Mathematics
1.Amazing Number Nine  2. Squaring A New WAY 3. Piyush Formula And Table For Squaring 4.Squaring Through Squares   5.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

 At Point No Of Points 1 7 13 19 25

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

If we do 1=1 + (1*0*…..)

2= (1+7) + (2*1*0) =8

3= (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)

N=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

 At Point No Of Points 1 5 9 13 17

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

If we do

1=1 – (1*0)

2= (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).

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