Buimama Anandita Theorem-Mathematics(Sri Chaitanya Techno School-Ameerpet)

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.