# Proof for nth term formula

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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

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

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#2

(Original post by

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

**aacharyauk**)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

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(Original post by

4th year Cambridge Maths Degree Level!!

**bono**)4th year Cambridge Maths Degree Level!!

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(Original post by

What does this formula represent?

**theone**)What does this formula represent?

d is the first difference in the sequence

c is the changing difference from one difference to the next

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look if anyone is out there and knows could u reply in the next few mins cos i am seriously desperate for a proof of this formula. It will be the end of two weeks worth of struggle.

PLZ PLZ PLZ PLZ

PLZ PLZ PLZ PLZ

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#8

(Original post by

a is the first term

d is the first difference in the sequence

c is the changing difference from one difference to the next

**aacharyauk**)a is the first term

d is the first difference in the sequence

c is the changing difference from one difference to the next

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

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#9

**aacharyauk**)

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

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.

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#10

(Original post by

This is called a geometric sequence:

**2776 2**)This is called a geometric sequence:

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

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#11

(Original post by

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.

**theone**)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.

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(Original post by

This is called a geometric sequence:

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

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

(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

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.

**2776 2**)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.

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#13

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).

Now let us consider your series, defined as:

a, a+d, a+2d+c, a+3d+2c ... a+(n-1)d+(n-2)c.

Now this is equal to:

a + a+d + ... a+(n-1)d = n/2(2a+(n-1)d). Also c + 2c + ... (n-2)c = (n-2)(n-1)c/2.

Adding gives an + n(n-1)d/2 + (n-2)(n-1)c/2.

This isn't the same as what you initally posted, but I think this is right. Hope this helps.

Is the series a, a+d, a+2d+c, a+3d+2c ... a+(n-1)d+(n-2)c. what you actually mean?

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).

Now let us consider your series, defined as:

a, a+d, a+2d+c, a+3d+2c ... a+(n-1)d+(n-2)c.

Now this is equal to:

a + a+d + ... a+(n-1)d = n/2(2a+(n-1)d). Also c + 2c + ... (n-2)c = (n-2)(n-1)c/2.

Adding gives an + n(n-1)d/2 + (n-2)(n-1)c/2.

This isn't the same as what you initally posted, but I think this is right. Hope this helps.

Is the series a, a+d, a+2d+c, a+3d+2c ... a+(n-1)d+(n-2)c. what you actually mean?

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(Original post by

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).

<editing>

**theone**)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).

<editing>

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#15

(Original post by

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.

**aacharyauk**)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.

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**aacharyauk**)

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.

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(Original post by

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

**aacharyauk**)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

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#19

(Original post by

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

**prince_capri**)is this thingy for A level or degree maths ???

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(Original post by

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

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

I used this formula to get the nth term of this series

1,3,6,10,15

here i had to look at the third difference to get a constant and hence the formula became

a + (n-1)d + (1/2)(n-1)(n-2)c1 + 1/6 (n-1)(n-2)(n-3)

and for this series

1, 5, 14, 30, 55

the formula remained as it was as at the second difference i had a constant.

Now i had to prove why this formula works for working out the nth term.

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