Math Is Fun Forum

1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + .............1/ ∞ = 2
I cannot think beyond that.

First, the formula or equation is checked for values like 1,2,3 etc.Only if this test is pased, we asume it is true for an arbitrary number k. Thereafter, if it is proved true for k+1, it becomes a 'proof beyond doubt'.

Mathematical Induction is a brilliant way of proving.
For example, the sum of the first n Natural Numbers is
n(n+1)/2.
For n=1, we get 1.
For n=2, we get 3.
Therefore, the summation formula works for 1 and 2.

Let's assume it is true for an arbitrary Natural Number k.
Therefore, the sum of the first k Natural Numbers would be k(k+1)/2.

For k+1, the sum would be k(k+1)/2 + (k+1),
that is [k(k+1) + 2(k+1)]/2, or [(k+1)(k+2)]/2

And, according to the formula we had at the beginning of the proof,
the sum of the first k+1 Natural Numbers would be [(k+1)(k+2)]/2

Since the two are equal, this can be said to be true for sum up to any natural number.

Similarly, it can be proved that:-
the sum 1² + 2² + 3² + 4² +... n²  = [n(n+1)(2n+1)]/6
and also
the sum 1³ + 2³  + 3³ + 4 ³  + ......... n³  = [n(n+1)/2]²

Hannah,
Mathematics is fun when you try to understand the subject.
Trying to memorize the tables and the formulae may be difficult for some.
Yet, it is easier to get a good score in Mathematics than any other subject.
For example, When you forget what 7x7 is,
try to remember what 7x6 is, then add 7 to the result.
You can get doubts clarified here.

Maybe, I should have waited for some more days
before giving the solution.

I am unable to find any serious flaw, yet,
I am not fully convinced.
I shall try with a much simpler number, viz. y76.
y76 is (y*10^2)+(7*10^1)+(6*10^0).
When this number is squared,
[This is of the form (a+b+c)²
which is equal to a² + b²  + c² + 2ab + 2bc + 2ac]
we get
10000y²  + 4900 + 36 + 14000y + 840 + 1200y
which is the same as
10000y² + 15200y + 5776
From this, it is clear, for any y belonging to Natural Numbers,
the last two digits are not affected. They continue to be 76.
But, does this conclusively prove there exists a value y such that
the last three digits of y76² are y76?

Mathsy, I am confused.

Mathsy, how did you get '2(y*(6*10^0))+a constant made up of the expansion of the other powers of 10, but that does not involve y'?

Both x! and x^y work, but for larger values, as you rightly pointed out, it becomes very slow.
However, a useful tool.

42576576769103890995893380022607743740081787109376

Mathsy, did you get this number too by exhaustion, just like you got the 30 digits????:)

I'll give my opinion tomorrow on Mathsy's proof, when I start the day afresh;
Regarding including n!, it would be of immense help as combinatorics works mostly on
factorials;
nPr= n!/(n-r)!
and
nCr = n!/(n-r)!*r!
nPr gives the number of Permutations
and
nCr gives the number of combinations.
There is an approximation for factorials of higher numbers;
it is called the James-Stirling formula
n! is approximately equal to
√(2*Pi*n) multiplied by (n/e)^n
This is a brilliant approximation as the value of n goes up.