The series of natural numbers is divided into groups (1),(2,3,4),(5,6,7,8,9)....and so on.
Find the sum of the numbers in the nth group.
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2 Answers
62the sum of the nth group is the sun of the first n2 terms minus the sum of first (n-1)2 terms
that is what?
106in 1st group there are 1 nos. 2nd grp - 3nos 3rg grp - 5nos
So, nth grp - (2n-1) nos
No. of terms before nth grp = 1 + 3 + 5 + .. + (2n-3) = (n-1)[2*1+(n-2)*2]/2 = (n-1)2
So, 1st term of nth grp + (n-1)2+1 and last term of nth grp = (n-1)2 + (2n-1)
So, sum of all terms in nth grp = (2n-1)(n-1)2 + (1+2+...+(2n-1))
= (2n-1)(n-1)2 + (2n-1)*2n/2
= (2n-1)(n2-n+1)
