First slide

Introduction to P.M.I

Question

The sum of the terms in the nth bracket of the series $(1) + (2 + 3 + 4) + (5 + 6 + 7 + 8 + 9) + \dots,$ is

Easy

A

$(n - 1)^{3} + n^{3}$
B

$(n + 1)^{3} + 8 n^{2}$
C

$\frac{(n + 1) (n + 2)}{6 n}$
D

$(n + 1)^{3} + n^{3}$

Solution

For $n = 1,$ we have
Sum of the terms in first bracket = 1
and, $(n - 1)^{3} + n^{3} = (1 - 1)^{3} + 1^{3} = 1$
For $n = 2,$ we have
3um of the terms in the second bracket $= 2 + 3 + 4 = 9$
and, $(n - 1)^{3} + n^{3} = (2 - 1)^{3} + 2^{3} = 1 + 8 = 9$
For $n = 3,$ we have
Sum of the terms in the third bracket $= (5 + 6 + 7 + 8 + 9) = 35$
and, $(3 - 1)^{3} + 3^{3} = 8 + 27 = 35$
Hence, option (a) is true.

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