Question

If 1(0!)+3.(1!)+7(2!)+13(3!)+21(4!)+⋯ n terms = (4000)(4000!) Then what is the value of n.

Answer 1

Answer:

n = 4000

Step-by-step explanation:

Sn = 1(0!)+3(1!)+7(2!)+13(3!)+...+tn.

tn = (n²-n+1).(n-1)!

= n².(n-1)! - (n-1).(n-1)!

= n.n! - (n-1).(n-1)!

Now,

1.(0!) = 1.1! - 0.0!

3.(1!) = 2.2! - 1.1!

7.(2!) = 3.3! - 2.2!

.

.

.

tn = n.n! - (n-1).(n-1)!

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Sum all terms, then

Sn = n.n!

Given, Sn = 4000(4000!) = n.n!

so,

n = 4000.