The series of natural numbers is divided into groups (1), (2, 3, 4), (5, 6, 7, 8, 9),and so on. The sum of first and last term of the nth group is

A
$2 (n^{2} - n + 1)$
B
$n^{2} - 2 n + 2$
C
$n^{2} + n$
D
$n^{2} - n + 2$

Solution:

Note that the nth group contain (2n – 1) natural number and the last term of the $n t h$ group is
$1 + 3 + 5 + \dots + (2 n - 1) = n^{2}$
Thus, sum of first and last term of $n t h$ group is
$n^{2} + (n - 1)^{2} + 1 = 2 (n^{2} - n + 1)$

The series of natural numbers is divided into groups $(1), (2, 3, 4), (5, 6, 7, 8, 9),$ and so on. The sum of first and last term of the $n t h$ group is

Solution:

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