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Tough Further Maths Question

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1. If Sn=n + 2(n-1) + 3(n-2)+...+(n-1)x2 +nx1, where n is a positive integer,
prove that Sn+1 - Sn=1/2(n+1)(n+2)
prove that Sn= 1/6(n)(n+1)(n+2)
2. (Original post by RNVS_1299)
If Sn=n + 2(n-1) + 3(n-2)+...+(n-1)x2 +nx1, where n is a positive integer,
prove that Sn+1 - Sn=1/2(n+1)(n+2)
prove that Sn= 1/6(n)(n+1)(n+2)
By (n-1)x2, do you mean or ? If it's the latter the sequence looks rather strange to me.

Also, what have you tried / what are your thoughts?
3. Hi,

For what you asked, it is 2(n-1).

Ive tried quite a few things,but I end up with a sequence that has r and n in it as variables which obviously wont work, so I Dont know how to go about doing this?!
4. (Original post by SeanFM)
By (n-1)x2, do you mean or ? If it's the latter the sequence looks rather strange to me.

Also, what have you tried / what are your thoughts?
Here is the question, badboy Sean - Q10
5. (Original post by RNVS_1299)
Hi,

For what you asked, it is 2(n-1).

Ive tried quite a few things,but I end up with a sequence that has r and n in it as variables which obviously wont work, so I Dont know how to go about doing this?!
(Original post by Chittesh14)
Here is the question, badboy Sean - Q10
Have spent some time on it but can't get my head around it. Sorry! Maybe someone else can have a go
6. (Original post by RNVS_1299)
If Sn=n + 2(n-1) + 3(n-2)+...+(n-1)x2 +nx1, where n is a positive integer,
prove that Sn+1 - Sn=1/2(n+1)(n+2)
prove that Sn= 1/6(n)(n+1)(n+2)
so .

there are n+1 terms here and n terms in so you can write on one line and on the line below like so;

http://imgur.com/a/QufU5

from there its fairly straightforward.

The next part is just proof by induction
7. (Original post by SeanFM)
Have spent some time on it but can't get my head around it. Sorry! Maybe someone else can have a go
Jheeze Sean, uni stress hitting u again!!!

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