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# Help!!!

0
354
6

I need:

an= n2-1

an= 2n3-2n

an= n2+n/2

Urgent!!!!!!!!

Nov 20, 2017

#1
+578
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no solution to this system sorry

Nov 20, 2017
#2
+96956
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We do not know what you are asking for. The question is incomplete.

Nov 20, 2017
#3
+1

I need:

an= n2-1

an= 2n3-2n

an= n2+n/2

Urgent!!!!!!!!

What do you want done with these? They appear to be formulas for the nth term of a sequence:

Example: Give the 10th term for each formula:

a(10) =10^2 -1 =99

a(10) =2*10^3 - 2*10 =1,980

a(10) =10^2 + 10/2=105

Nov 20, 2017
edited by Guest  Nov 20, 2017
#4
+1

Ok, if it is true is that I forget to put everything, I need to find the first six terms of each sequence, but I really do not understand how it is done, if something should be replaced or not

Guest Nov 20, 2017
#5
+1

Well, just sub 1, 2, 3, 4, 5, 6  for n in each formula, and that will give you the first 6 terms:

a(n) =n^2 - 1

a(1) =1^2 - 1 =1 - 1 =0 that is your first term.

a(2) =2^2 -1 =4 - 1 =3 that is your second term

a(3) =3^2 - 1 =9 -1 =8 that is your third term

a(4) =4^2 - 1 =16 -1=15 that is your fourth term

a(5) =5^2 -1 =25 - 1 =24 that is your fifth term

a(6) =6^2 - 1 =36 -1 =35 and that is your sixth term.

Do exactly the same in the second formula:

a(n) =2*n^3 - 2*n, sub 1, 2, 3, 4, 5, 6 for n

a(1) =2* 1^3 - 2* 1=2 - 2 =0 your 1st. term

a(2) =2* 2^3 - 2*2=16 - 4 =12 your 2nd term.

a(3) =2* 3^3 - 2*3 =54 - 6 =48 your 3rd term

a(4) =2*4^3 - 2*4 =128 - 8 =120 your 4th term

a(5) =2* 5^3 - 2*5 =250 - 10 =240 your 5th term

a(6) =2* 6^3 - 2*6 =432 - 12 =420 your 6th term

Do exactly the same in the third formula:

a(n) =n^2 + n/2 sub 1, 2, 3, 4, 5, 6 for n

a(1) =1^2 + 1/2 =1 1/2 your 1st term

a(2) =2^2 + 2/2 =4 + 1 =5 your 2nd term

a(3) =3^2 + 3/2 =9 + 1 1/2 =10 1/2 your 3rd term

a(4) =4^2 + 4/2 =16 + 2 =18 your 4th term

a(5) =5^2 + 5/2 =25 + 2 1/2 =27 1/2 your 5th term

a(6) =6^2 + 6/2 =36 + 3 =39 your 6th term.

AND THAT IS IT !!. DO YOU UNDERSTAND IT??. CHECK MY MATH TO MAKE SURE I DIDN'T MAKE ANY MISTAKES.

Nov 20, 2017
#6
+1

thank you very much!!! it was very helpful

Guest Nov 20, 2017