I need a simpler formula for that expression. Im supposed to prove by induction that the number of squares within an n x n checkerboard is that expression. I definately need a simpler formula, but any help towards the proof would be appreciated, thanks!
The formula for the sum of squares of first n natural numbers is given by:
1^2+2^2+3^2+..........+n^2 = n(n+1)(2n+1)/6, ................ eqn(1)
and as far as the proof is concerned, it can be proved by using the concept of mathematical induction as follows:-
Let n= 1, therefore L.H.S of the above expression becomes
1^2 = 1
and R.H.S becomes
Hence the result is true for n=1.
Let us suppose that the result is true for n=k,i.e.
1^2+2^2+3^2+..........+k^2 = k(k+1)(2k+1)/6
Let us prove the result for n=k+1.
If we put n = k+1 on the L.H.S of the eqn (1) then it becomes
=k(k+1)(2k+1)/6 + (k+1)^2
Now let us put n=k+1 on the R.H.S of eqn (1),we have
Hence L.H.S = R.H.S,
Hence the result is true for n=k+1 whenever it is true for n=k.
Thus it is true for all n.
I hope this will help you in clarifying your doubts.
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In general, you can easily estimate
But before let me introduce some basic concepts on difference operator, factorial function and the Stirling numbers.
The forwards difference operator is defined to be .
And the factorial function for is defined as (exactly numbers of successive terms are being factored), and for convenience .
It is not hard to see that
Now, let and satisfy the following relations:
which are called as the Stirling numbers of first and second kind, respectively (also see Stirling number - Wikipedia, the free encyclopedia).
Then, lets turn back to the problem.
Finally, if is known, you may simplify the last term above.
Check the result for .
Judging from the number of off-topic posts this thread collected at its tail-end (and which you can now find here: http://www.mathhelpforum.com/math-he...countries.html), I'd say that nothing much more that's on-topic is going to get added so (drum roll, maestro, if you please) .......