Alan

This should help:

You are right!  But I'll leave it to someone else to include that!

Here's a simpler way!

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As follows:

(Can't help feeling there's probably a simpler way!)

Here's an alternative approach:

You could try using WolframAlpha to factor your large integers. For example:

"can you explain why you did Ur + U n+1 at the first part of ii please "

Think of the summation as follows:  \(\sum_{k=1}^nu_k=u_1+u_2+...+u_n\)

Hence \(\sum_{k=1}^{n+1}u_k=u_1+u_2+...+u_n+u_{n+1}\)  which is \(\sum_{k=1}^{n+1}u_k=\sum_{k=1}^nu_k+u_{n+1}\)

Hope that makes it clear.

Solution for part b:

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Here's my go at this (I might also have made a careless error!):

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There are 4 values of n:

n           S(n)       S(S(n))

1977      24           6

1980      18           9

1983      21           3

2001        3           3