Proof for nth term formula

Im really desperate
does anyone know the proof for the following nth term formula when the difference is changing.

a(n-1)d+1/2(n-1)(n-2)c

This is a formula from the cgp higher maths GCSE book-but i need to find a proof to show why it works. If anyone knows a proof or can work one out could they plz plz post it clearly in stages.

Thanks

aaaarghhhhhh maths!!

4th year Cambridge Maths Degree Level!!

What does this formula represent?

a is the first term
d is the first difference in the sequence
c is the changing difference from one difference to the next

Im really desperate
does anyone know the proof for the following nth term formula when the difference is changing.

a(n-1)d+1/2(n-1)(n-2)c

This is a formula from the cgp higher maths GCSE book-but i need to find a proof to show why it works. If anyone knows a proof or can work one out could they plz plz post it clearly in stages.

Thanks

This is called a geometric sequence:

Actually this is not a geometric series since the difference between terms does not increase exponentially.

For instance the series 1,2,4,7,11,16 has a=1, b=1 and c=1 but is not geometric.

This is called a geometric sequence:

Let S, be the sum of a sequence...

(S = 1 + 2 + 4 + 8 + 16.)

Times this by 2:

(2S = 2 + 4 + 8 + 16 + 32)

Notice that both have 2, 4, 8, 16 in common, and the sum of this is S-1 rom teh first equation, and 2S-32 from the second. So:

S - 1 = 2S - 32, giving S=31

This can be used to find the sum of any geomeric sequence. Let S be the sum of the first n term of the sequence. Then:

S = a + ar + ar^2 + ... + ar^n-2 + ar^n-1.

If you multiply is by r, you get:

Sr = ar + ar^2 + + ar^3 + ... ar^n-1 + ar^n

The right sides in these 2 equations have the terms ar + ar^2 + ... ar^n-2 + ar^n-1 in common, so:

S-a = ar + ar^2 + ... ar^n-2 + ar^n-1 = Sr - ar^n

Which gives:

S(1-r) = a(1-r^n)

And so S = [a(1-r^n)]/[1-r]

Notice that r can both be positive and negative for this to work.

First let us consider the simpler series:

a, a+b, a+2b... a+(n-1)b. Let the sum of this be S. Now writing this backwards, S = a+(n-1)b + ... a+b + a. Now adding corresponding terms. 2S = n(2a + (n-1)b) so S = n/2(2a+(n-1)b).

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what do u mean-i am really thick at maths. Sorry but seriously all i am going to do is copy your proof. Ill try and understand it tho.

what do u mean-i am really thick at maths. Sorry but seriously all i am going to do is copy your proof. Ill try and understand it tho.

U see the formula i posted is the one for working out the nth term. Now i need to put this formula in my work but cos its from a book i cant just put it in. I need to prove it and say why it works. Algebraically

is this thingy for A level or degree maths ???

Can you explain what kind of series this is? And i mean explain it properly.