Nth Term of a Quadratic Sequence - difference method
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@armaan3495 nope, that works too
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@whatheduck97 You then do the sequence take away the 4n*2 terms. so it would be 2-4 = -2, 17-16 = 1, 40-36 = 4 etc., the numbers you get make a linear equation and they go up by 3 each time so the "bn" term is 3n. to get the "c" term you go back one space so you subtract three from the first term which is -2. so -2 - 3=-5. so the "c" term is -5. you then bring all the "an*2, bn and c" terms together to give 4n*2+3n-5. hope this helped.
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@whatheduck97 a quadratic sequence will always look like an*2 + bn + c e.g. 5n*2+4n+3. to get the a term you just divide the 2nd difference by two. in this question the second difference is 8, so you divide 8 by 2 to get 4. this then acts as the a term so it is 4n*2. you then need to get the "bn+c" term so you would do the sequence(2,17,40,71 etc.) minus 4n*2. you would input the normal numbers into 4n*2 e.g. 4(1)*2 = 4, 4(2)*2 = 16, 4(3)*2 = 36 etc.
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you just increased my chances of getting an A in my igcse exam tomorrow :D thanks!!
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would it be wrong if i find the value of 'a' and then find 'b' & 'c' by solving two equations simultaneously?
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erm i hope someone will reply, i understand it all up to 1:16, and then i just lose it, can someone please explain it in a really simple way?
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Why did you use this program? - it completely confused me affter you found out "A"...
Bane391 8 months ago 5