The series of natural numbers is divided into groups

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The series of natural numbers is divided into groups as follows ; (1), (2, 3), (4, 5, 6),

(7, 8, 9, 10) and so on. The sum of the numbers in the nth group is(A) 1/2{n(n° +1)](B) n(n^2+1)/4(C) 2n(n + 1)/3(D) n^2(n+1) /2​

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Answer:Description for Correct answer:

Let S =1+2+4+7+...+Tn

or S = 1+2+4+...+Tn−1+Tn

Subtracting, we get

O = 1+[1+2+3+...(n-1)]-Tn

=> Tn = 1+2+3+.....+(n-1)+1

= n(n−1)2+1

∴First number of 50th term

= 50×492+1= 1226

∴Sum of numbers of 50th term

= 1226 + 1227 + ......upto 50th

term =502[2×1226+(50−1)×1]=25×2501=62525

Explanation:


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