This is a very simple mathematical relationship I found out about recently, which I rather like (although I don’t claim to be original… it’s probably been covered millions of times before!). Here’s an example:

1^{2} = 1

2^{2} = 4 = 1 + 3

3^{2} = 8 = 1 + 3 + 5

4^{2} = 16 = 1 + 3 + 5 + 7

5^{2} = 25 = 1 + 3 + 5 + 7 + 9

You’ll notice that the square numbers are being formed by sums of consecutive odd numbers, starting at 1. I don’t have a mathematical proof for this (although I’m sure one exists), but it appears to work for any *even* power. I wrote a quick computer program to try the same with the 10^{th} power, and sure enough, it works:

1^{10} = 1

2^{10} = 1024 = (first 32 odd numbers)

3^{10} = 59049 = (first 243 odd numbers)

4^{10} = 1048576 = (first 1024 odd numbers)

5^{10} = 9765625 = (first 3125 odd numbers)

Do you notice the curious relationship between the raised base and the number of summed odd numbers required to reach it? Here’s another way to write it:

1^{2} = 1

32^{2} = 1024

243^{2} = 59049

1024^{2} = 1048576

3215^{2} = 9765625

So the raised base is the square of the number of summed odd numbers required to reach it. In other words, where “y” is any positive even number:

(You may recognise “2n – 1” as being the way you calculate the n^{th} odd number starting from 1.)

We can actually simplify it a little however. Notice that we’re taking the square root of a power? A root is actually just a fraction of a power, i.e. a square root is the power 1/2, and a cube root is the power 1/3. As such, we can simplify this by multiplying the powers together, which means simply dividing the power *y* in half on our summation. But now we have uncovered a (very bizarrely convoluted) demonstration of why this doesn’t work for all odd powers — if you divide an odd number in half, you get a fraction of a number, so (generally speaking) you will end up trying to carry the summation on to an non-integer limit, which is not possible. (Worth noting that it does work with odd powers where the base is itself a square number, since that allows the lingering square root in the power to be kind of ‘cancelled’ out, giving an integer in the end.)

Anyway, here’s a slightly simplified form of the equation, where “y” is any positive integer:

(You could also write the exponent over the summation as just “y/2”, but this communicates a little better that our final exponent ought to be even.)

Now, I’m sure you’re wondering why on earth that is useful, given that you still need to figure out a power in order to calculate how many odd numbers you need to add up in order to calculate your first power. And you are probably right… it’s not much good… *unless* the exponent you are dealing with is a power of 2… in which case you could conceivably perform some nested summations, where you calculate the upper limit of a given summation using another summation, and so on, all the way from 1 up to half of your original exponent.

(Only using powers-of-two exponents might seem like a limitation, but I’m sure anybody who’s done a good deal of programming or other computery stuff will prick up their ears at that.)

The nested summation sounds hideously complicated though! ðŸ˜€ That will be a challenge for another day perhaps. Or at least, when I’ve had more caffeine.

I was reading a history of Mathematics, in which I learned that Nichomachus of Gerasa observed that cubical numbers are the sum of successive odd numbers. The example given in the book was 8 = 2^3 = 3+5, 3^3 = 27 = 7+9+11, 4^3 = 64 = 13+15+17+19.

I found it curious that in each of the examples, the number of odd numbers equaled the number being raised to the third power. So I delved into it and, to make a long story short, was able to prove that this is true in general for any power, namely that

for n and r Natural numbers, both > 0:

n^r = Sum, for i = 0 to n-1, n^(r-1) – n + 1 + 2*i

So for fourth powers, for example, 7^4 = Sum, for i = 0 to 6, 7^3 – 7 + 1 + 2*i

= Sum, for i = 0 to 6, 343 – 7 + 1 + 2*i

= Sum, for i = 0 to 6, 337 + 2*i

= 337 + 339 + 341 + 343 + 345 + 347 + 349

= 2401, which is indeed 7^4

Interestingly, for squares, the formula just reduces to the sum of the first n odd numbers.

How one will considere the power of a no. Whether it is odd or even ex. 4 raise to power 981 ?

I think you have to divide the power by four and you will get the power of number as odd or even, as in this case it is odd.