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Worked solution

Recognise factorial growth from the recurrence

Find the first seven terms of un+1=(n+1)unu_{n+1}=(n+1)u_n, u1=1u_1=1, and a formula for unu_n.

CJC · Review problems · Question 12

Answer

1,2,6,24,120,720,50401,2,6,24,120,720,5040; un=n!u_n=n!.

Generate the terms using the immediately preceding term

Successive multiplication gives:

nu_n
11
22
36
424
5120
6720
75040

For example u5=5u4=120u_5=5u_4=120, u6=6u5=720u_6=6u_5=720, u7=7u6=5040u_7=7u_6=5040. Some source symbolic lines insert stray +1+1 terms and use u3u_3 in the u5u_5 line; the stated recurrence, numerical products and calculator table consistently give the values above.

How did this part go?

Read the general product and the calculator pattern

un=n(n−1)⋯2⋅1=n!u_n=n(n-1)\cdots2\cdot1=n!. The source also obtains the same first seven values from a calculator recurrence table and recognises 1!,2!,…,7!1!,2!,\ldots,7!. Induction confirms the pattern: from un=n!u_n=n!, un+1=(n+1)n!=(n+1)!u_{n+1}=(n+1)n!=(n+1)!, with base u1=1!u_1=1!.

How did this part go?

Common mistakes

Use the stated recurrence rather than the stray plus-one symbols in some printed working lines.

Read the hint