KVPY SB SX Exam 2011 Question Paper

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Question : 19 of 120

Marks: +1, -0

The sum of

(1^2 - 1 + 1)(1!) +

(2^2-2 + 1)(2 !) +

... + (n^2-n+1)(n!)

is

$(n + 2)!$
$(n-1)((n+1)!)+1$
$(n + 2)! - 1$
$n((n+1)!)-1$

Solution:

T_n = (n^2 - n + 1) n!

= (n^2 - 1) n! - (n - 2) n!

T_n = (n - 1) (n + 1) ! - (n -2) n!

Sum

= 1 + (n - 1) (n + 1) !

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