Question

1.
Let n! = 1 x 2 x 3 x 4 x .... x n for integer n 2 1. What could be the value of 1! + (2 x 2!)
+ (3 x 3!) + (4 x 4!) + ... (10 x 10!)?
1) 11X11!
2) 11!
3) 11! - 1
4) 11! + 1​

Answer 1

Answer:-Answer: (d)

If P = 1! = 1

Then P + 2 = 3, when divided by 2! remainder will be 1.

If P = 1! + 2 × 2! = 5

Then, P + 2 = 7 when divided by 3! remainder is still 1.

Hence, P = 1! + (2 × 2!) + (3 × 3!)+ ……+ (10 × 10!)

Hence, when p + 2 is divided by 11!, the remainder is 1.

Alternative method:

P = 1 + 2 × 2! + 3 × 3! + …..10 × 10!

= (2 – 1)1! + (3 – 1)2! + (4 – 1)3! + …+ (11 – 1)10!

= 2! – 1! + 3! – 2! + …..+ 11! – 10!

= 11! – 1

Hence p + 2 = 11! + 1

Hence, when p + 2 is divided by 11!, the remainder is 1

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Answer 2

Answer:

3) 11!-1

Step-by-step explanation:

P = 1 + 2 × 2! + 3 × 3! + …..10 × 10!

= (2 – 1)1! + (3 – 1)2! + (4 – 1)3! + …+ (11 – 1)10!

= 2! – 1! + 3! – 2! + …..+ 11! – 10!

= 11! – 1