28 Oct 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.
- We take 3 Pyramid faces and mark the Left Side with 1, 2, 3, 4,…….so on.
- 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 N^{3 }=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^{3 }=1 + (1*0*…..)
2^{3 }= (1+7) + (2*1*0) =8
3^{3 }= (1+7+13) + (3*2*1) =21 + 6 = 27 and so on
N^{3}=(N/2)[2a+(N-1)*d] + C(N,3)
Putting the value a=1 & d=6, we get
N^{3}=(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^{3 }=N (3N-2) + N (N-1) (N-2) (Hence Proved)
Secondly, we are finding Square of a number.
- We take 2 Pyramid faces and mark the Left Side with 1, 2, 3, 4,…….so on.
- 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 N^{2}=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^{2 }=1 – (1*0)
2^{2 }= (1+5) – (2*1) = 6 – 2 =4
3^{2}= (1+5+9) – (3*2) =15 – 6 = 9
4^{2}= (1+5+9+13) – (4*3) =28 – 12 = 16
N^{2}=(N/2)[2a+(N-1)*d] – C(N,2)
Putting the value a=1 & d=4, we get
N^{2}=(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, N^{2}=N (2N-1) –N (N-1).
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