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Question

The sequence of natural numbers is divided into classes as follows
Prove that the sum of the numbers in the nth row is n(2n2+1)
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Solution

The number of numbers in successive rows are 2, 4, 6 and hence there will be 2n numbers in nth row which will be in A.P. of common difference 1. The first term in successive rows are
S = 1 + 3 + 7 + 11......+ tn
S = 1 + 3 + 7 + 11......+ tn1tn
Subtracting,
0 = 1 + [2 + 4 + 6 + (n - 1) terms] - tn
tn=1+2(n1)n2=n2n+1
Sn = sum of an A.P in which
A = n2 - n + 1, D = 1, N = 2n.
Sn=2π2[2(n2n+1)(2n1)1]
=n(2n2+1)

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