__Creativethoughts-12__We know the sum of n numbers from 1 in the natural series is n(n+1)/2

S = 1+2+3+4 +5+6+7+8+………. n = n(n+1)/2

If n is an even number, then the first and last numbers, second and the last but one

numbers of the series ……… added together each sum will be equal to (n+1).

As there are n/2 such pairs, the total sum will be (n/2)(n+1). If n is odd, we get the

same result.

Here all the number can be paired up except the central number i.e., (n+1)/2,

the number of pairs will be (n-1)/2 and the sum of each pair is (n+1) . Hence its sum will be

S = [(n-1)/2][n+1] + (n+1)/2 = n(n+1)/2

It predicts that for twice the sum of n numbers from 1 in natural series , n and (n+1)

are factors. With numerical examples,

1+2 = 3 1 x3

1+2+3 = 6 = 2 x3

1+2+3+4 = 10 = 2x5

1+2+3+4+5 = 15 = 3 x 5

1+2+3+4+5+6 = 21 = 3 x 7

1+2+3+4+5+6+7 = 28 = 4x 7

I f the number of numbers added together is even ,then its sum can be represented

as a product of two numbers (n/2)and (n+1) ,they are n and [(n+1)/2],if it is odd.

The sum of numbers in natural series from smaller number n(s) to higher number n(h)

S = n(s) + [n(s)+1] + [(n(s)+2] + ………… n(h)

n(h) = n(s) + N-1,where N is the number of numbers in the series

The sum of numbers in natural series from 1 to n(h) is [(n(h)][(n(h)+1)]/2 and the

sum of number in natural series from 1 to n(s) – 1 is [n(s)-1][n(s)]/2

The difference between these two sums gives the required sum and is given by,

S = S(h) – S(L-1) = [n(h)][(n(h)+1)/2] – [n(s)-1][n(s)]/2

=[ n(h)xn(h) + n(h) – n(s)xn(s) + n(s)]/2

{[n(h)-n(s)][n(h)+n(s)] + [n(h) + n(s)]}/2

= [n(h)+n(s)][n(h)-n(s)+1]/2

where n(h) + n(s) is the sum of the initial and final terms of the series and

n(h)-n(s) + 1 denotes the number of terms added up in the given series.

It opens an avenue for quite a large number of mathematical puzzles. The sum

of a set of n numbers in natural series is S ,find the possible natural series with

same sum but with different number of terms. We know,

Twice of the sum = (sum of the first and last terms) x (number of terms in the series)

= [n(s) + n(h) ] N

If the twice of the sum is divisible by N-x , then there will be one or more another

solutions. For example,

S = 11+12+13+14+15+16 = 81

2S = 162 which is divisible by 6, the number of terms in the given series.

162/6 = 27 = n(s)+n(h) .It is not divisible by 5,4 but by 3.

162/3 = 54 = n(s) + n(h)

so there must be a series with three terms whose first and last terms give 54 on addition.

S = 26 + 27 + 28= 81

It is divisible by2 as well, so

S = 40+41 = 81

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