Introduction: Anandita Theorm:
Anandita : This theorem talks about an alternate formula of Gauss Theorem of
addition of (n) WHOLE numbers
Reference:
Gauss Theorem:
Sum of first N Natural Numbers{N} like 1+2+3+4+5+6+....n = n(n+1)/2
As
per Anandita's theorem:
Sum of first N Whole Numbers STARTING WTH 0 (Whole Numbers{W}
= 0 + {N}) = n(n-1)/2
Where
n = Number of digits in Whole Number Series including 0.(i.e as per
Gauss Theorem n = n + 1)
Proof...
As per Gauss Theorem: 1+2+3+4 = 4x(4+1)/2 = 4x5/2 = 10 (Using
nx(n+1)/2 theorem)
But,
Sum Of Whole Numbers(Means 0 is included in that Series of Sum, so
number of terms n changes, i.e it becomes n+1)
=>
0+1+2+3+4 = 5x(5-1)/2= 5x4/2 = 10 (Here n= number of terms = 5
including 0)
So,
the Sum is same as nx(n+1)/2.
I
didn't understand first,but in her Theorem , 0 is added as a number in
the series, because when 0 is added, the series becomes a Whole
Number Series.
As
its a sum of Whole Numbers, where 0 is added.
So
when 0 is added, Sum value is un-altered. But Gauss theorem is not
applicable there, as Number of terms changes.
nx(n+1)/2
becomes nx(n-1)/2 when 0 is added to Gauss Theorem series.
where
n = Number of Natural Numbers + 1(i.e 0 which is a Whole Number)
Note:
0 is not a Natural Number{N}, but a whole Number{W}
I
tried a lot to prove the theorem wrong.But I failed..Simple logic is,
0 is added, where Sum is not altered, but Number of terms altered and
a
polynomial equation of nx(n+1)/2 and nx(n-1)/2 remains same, when n =
n+1
This
is main(Anandita doesn't know abt Polynomials yet in detail, but true)
A
solution of 2 polynomial equations i.e nx(n+1)/2 and nx(n-1)/2
remains same, when n = n+1
if
we replace n with (n-1) in Gauss Theorem, we reach there, but at least
the statement Sum of first N Whole Numbers STARTING WTH 0 (Whole
Numbers{W} = 0 + {N}) = n(n-1)/2 doesn't exist in Maths world yet.