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Let us assume that ’n' is even, hence there are equal numbers of positive and negative signed digits.
Therefore 1-3+5-7+9-11... Upto n terms can also be expressed as (1+3+5+7+……upto n terms) -2(3+7+11+…upto n/2
terms) . The 1st part is a A. p having common difference 2 and the first term 1. The 2nd part is a A. P with c. d =4 and first term =3. Hence by calculating it using summation formula we get the the value (-n) when ’n' is even. In case of ’n' being a odd number we can express the sum as (1+3+5+7+9……upto n terms) -2(3+7+11+15+….upto (n-1) /2 terms). As the series always end with a +ve signed digit. Therfore the number of +ve signed digits exceeds the number of -ve signed digits by 1. Again using summation formula we find the value to be equal to +n. Therefore the required sum is +n (when n is a odd number) or (-n) ( when n is even)
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