Maths HELP!?
Use a difference table to work out the rule for the Nth Term of these sequences:
a) 10,40,90,160
b)4,7,12,19
g)3,9,19,33,51
j)4,11,22,37,56
m)3,16,39,72,115
Answers:
here's one method of doing this problem
a) 10,40,90,160
1st diff >>> 30,50,70 2nd diff >>>>20,20
this means that the formula for N is a second order polynomial
of the form A n^2 +Bn+C....(1)
we need to form three simultaneous equations to find A, Band C
we will use the first three terms -sub into (1)
n=1, 10=A+B+C
n=2, 40=4A+2B+C
n=3, 90=9A+3B+C
solving gives A=10, B=0 and C=0
hence, N=10n^2 (n=1,2.....)
b) 4,7 12,19
here again we have a second difference and a polynomial
of the form A^2+Bn+C
using the same method as in a) we found that
A=1,B=0 and C=3
hence, N=n^2 +3 (n=1,2......)
g) 3,9,19,33,51 here again we found a second difference
and found A=2,B=0 and C=1
hence, N=2n^2 +1 (n=1,2,.......)
j) 4,11,22,37,56 here again we found a second difference
and found A=2,B=1 and C=1
hence, N= 2n^2 +n +1 (n=1,2,.....)
m) 3,16,39,72,115 here again we found a second difference
and found A=5,B= -2 and C=0
hence, N= n(5n-2) (n=1,2,......)
i hope that this helps
a)d1 = 30 50 70
d2= 20 20 20 so diff increases by 20 each time.next diff will be 90, 160 + 90 = 270
b) d1 = 3,5,7, sam principal as above diff inc by 2 next diff will be 9 ,19 +9 = 28
g) d1 = 6,10,14,18, diff goes up by 4,next diff willk be 22, 51+22=73
j)d1=7,11,15,19 going up by 4 again,next is23,56+23=78
m)d1=13,23,33,43,inc by 10,next is 53,115+53=168
a) 40-10=30, 90-40=50, 160-90=70
So a_n+1 - a_n = (2n+1)*10, where a0 is the first term in the sequence, 10.
Therefore, a_n = 10 + sum i=1 to n 10*(2i+1)
= 10 + 10 sum i=1 to n (2i+1)
= 10 + 20 sum i=1 to n (i) + 10n
= 10 + 10n + 10n(n+1), since the sum of i from 1 to n is n(n+1)/2.
b) very similar, just that the steps are odd numbers, instead of odd multiples of 10
g) differences are 6, 10, 14, etc. so they have the form 2+4n
j) very similar to g
m) differences are 3+10n
Once you have the formula for the differences, follow what I did in (a) to get the formula for the nth term, they're all similar in that respect
Go to http://labs.google.com/sets . It allows you to type in a set and then continues the trend for you.
a) 10 n^2
b) n^2 + 3
g) 2 n^2 + 1
j) 2 n^2 + n + 1
m) 5 n^2 - 2 n
250
28
73
168
The next sequence of numbers is...
a) 250
b) 26
g) 73
j) 79
m) 168
...you do the working out for the Nth Term...I'm not your mom...?
A difference table is just that - a table of differences. If you develope one for each series you'll see a pattern emerge.
For example:
a) 10, 40, 90, 160
Start with the first 2 integers in the series (10 and 40) and find their difference:
40 - 10 = 30
Now find the difference between 40 and 90:
90 - 40 = 50
Now find the difference between 90 and 160:
160 - 90 = 70
From this we can see that each difference is an odd placement of 10's, or more precisely, we can say that the difference between each term is found by 10(2n+1), as 2n+1 is the odd series.
The same approach can be used to find the rest of them, but they may involve different series formulas. It will be easy to check in your appendix, or extrapolate the series, once you've found the general pattern.
a) 10 x n^2
b) (n^2) + 3
g) 2(n^2) +1
j) 2(n^2) +n + 1
m) 5(n^2) - 2n
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a) 10,40,90,160
b)4,7,12,19
g)3,9,19,33,51
j)4,11,22,37,56
m)3,16,39,72,115
Answers:
here's one method of doing this problem
a) 10,40,90,160
1st diff >>> 30,50,70 2nd diff >>>>20,20
this means that the formula for N is a second order polynomial
of the form A n^2 +Bn+C....(1)
we need to form three simultaneous equations to find A, Band C
we will use the first three terms -sub into (1)
n=1, 10=A+B+C
n=2, 40=4A+2B+C
n=3, 90=9A+3B+C
solving gives A=10, B=0 and C=0
hence, N=10n^2 (n=1,2.....)
b) 4,7 12,19
here again we have a second difference and a polynomial
of the form A^2+Bn+C
using the same method as in a) we found that
A=1,B=0 and C=3
hence, N=n^2 +3 (n=1,2......)
g) 3,9,19,33,51 here again we found a second difference
and found A=2,B=0 and C=1
hence, N=2n^2 +1 (n=1,2,.......)
j) 4,11,22,37,56 here again we found a second difference
and found A=2,B=1 and C=1
hence, N= 2n^2 +n +1 (n=1,2,.....)
m) 3,16,39,72,115 here again we found a second difference
and found A=5,B= -2 and C=0
hence, N= n(5n-2) (n=1,2,......)
i hope that this helps
a)d1 = 30 50 70
d2= 20 20 20 so diff increases by 20 each time.next diff will be 90, 160 + 90 = 270
b) d1 = 3,5,7, sam principal as above diff inc by 2 next diff will be 9 ,19 +9 = 28
g) d1 = 6,10,14,18, diff goes up by 4,next diff willk be 22, 51+22=73
j)d1=7,11,15,19 going up by 4 again,next is23,56+23=78
m)d1=13,23,33,43,inc by 10,next is 53,115+53=168
a) 40-10=30, 90-40=50, 160-90=70
So a_n+1 - a_n = (2n+1)*10, where a0 is the first term in the sequence, 10.
Therefore, a_n = 10 + sum i=1 to n 10*(2i+1)
= 10 + 10 sum i=1 to n (2i+1)
= 10 + 20 sum i=1 to n (i) + 10n
= 10 + 10n + 10n(n+1), since the sum of i from 1 to n is n(n+1)/2.
b) very similar, just that the steps are odd numbers, instead of odd multiples of 10
g) differences are 6, 10, 14, etc. so they have the form 2+4n
j) very similar to g
m) differences are 3+10n
Once you have the formula for the differences, follow what I did in (a) to get the formula for the nth term, they're all similar in that respect
Go to http://labs.google.com/sets . It allows you to type in a set and then continues the trend for you.
a) 10 n^2
b) n^2 + 3
g) 2 n^2 + 1
j) 2 n^2 + n + 1
m) 5 n^2 - 2 n
250
28
73
168
The next sequence of numbers is...
a) 250
b) 26
g) 73
j) 79
m) 168
...you do the working out for the Nth Term...I'm not your mom...?
A difference table is just that - a table of differences. If you develope one for each series you'll see a pattern emerge.
For example:
a) 10, 40, 90, 160
Start with the first 2 integers in the series (10 and 40) and find their difference:
40 - 10 = 30
Now find the difference between 40 and 90:
90 - 40 = 50
Now find the difference between 90 and 160:
160 - 90 = 70
From this we can see that each difference is an odd placement of 10's, or more precisely, we can say that the difference between each term is found by 10(2n+1), as 2n+1 is the odd series.
The same approach can be used to find the rest of them, but they may involve different series formulas. It will be easy to check in your appendix, or extrapolate the series, once you've found the general pattern.
a) 10 x n^2
b) (n^2) + 3
g) 2(n^2) +1
j) 2(n^2) +n + 1
m) 5(n^2) - 2n
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