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:

1(0!)+3(1!)+7(2!)+13(3!)+21(4!)+. . . n terms, then

tn = (n²-n+1)(n-1)!

= [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!

13(3!) = 4.4! - 3.3!

.

.

.

tn-1 = (n-1).(n-1)! - (n-2).(n-2)!

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

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Sn = n.n!

so,

1(0!)+3(1!)+7(2!)+. . . n terms = 4000(4000!)

n.n! = 4000(4000!)

By comparison,

n = 4000.