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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

Scroll to see replies

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

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!!

rednirt

aaaarghhhhhh maths!!

4th year Cambridge Maths Degree Level!!

bono

4th year Cambridge Maths Degree Level!!

bono-plz plz what do u mean-plz reply back quickly. Can u give me a proof. What do u mean by cambridge-i need is desperately

theone

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

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

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

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 ???

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

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:

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.

2 2776

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.

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.

Is this proof for the formula a(n-1)d+1/2(n-1)(n-2)c- cos i am doing GCSE maths coursework for squares in rectangles and the thing is that i have found an algebraic formula but i have used this formula to get it. Now this formula is from a textbook and is not my work. My teacher said i have to prove this. So if anyone knows a proof for this formula-ie in words or algebra(preferably) of why it gives you the nth term then i can include this in my coursework to get higher marks. I need it quickly cos i gotta hand it in tommorrow. THANKS u guys for trying to help. I appreciate it.

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?

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>

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>

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.

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

I know- i never copy anyones work but im in a real panic that why! Ill try and understand it tho.

theone

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

OK -im really sorry. Right ummmmm

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.

- The nth term of a sequence is n^2+n
- Arithmetic series- sum of first ten terms is 460
- Question about Un = Sn - S(n-1) formula (Sequences and series)
- Do you think A Level Maths has enough proof in it?
- Proving a term is in a quadratic sequence
- Arithmetic sequence diverge or converge formula?
- Arithmetic or geometric sequence
- How do I use proof by induction on this, specifically using f(k+1)-f(k)?
- Sequences
- Maths - bouncing ball question
- a level maths first principle question
- Help with question
- A-level Edexcel Maths - Proofs
- gcse maths topic
- Math Proof Questions
- Advanced mathematics
- MAT practice
- Maths A level - Trignometric proof
- Probability Question
- Connection between KEavg and KEtotal

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