1. ## cubic sequence

hello i have been asked to use sigma notation.

i have a sequence here : 7,25,63,129,231,377,575.

aparantly i can use a sigma notation for this.

this is for a study which i am carrying out. above are sequences for total number of 3d squares for each sequences.

i got a formula of 2d already which is 2n^2-2n+1 and i also got a formula for 3d using the formula an^3+bn^2+cn+d.

but as a further part of the study i have to carry out sigma notation.

i have attached the 2d and 3d pictures. for 2d you will reliase i have black squares please ignore that as that is a sepate part of the study. as the moment i am looking at the total number of squares.

all help apreichated

thank you.

2. I think this is what's going on here. You're looking at the number of cubes in a sequence of figures of which an example is shown. Let's look at the number of cubes in each layer: call that L(n). Each layer is composed of a number of rows each of which has an odd number of cubes in it. So L(n) = 1 + 3 + 5 + ... + (2n-1) + (2n+1) + (2n-1) + ... + 5 + 3 + 1 = sigma_{i=0}^{n} (2i+1) + sigma_{i=0}^{n-1} (2i+1). Now sigma_{i=0}^n i = n(n+1)/2 so sigma_{i=0}^n 2i+1 = n(n+1) + n+1 = (n+1)^2. So L(n) = (n+1)^2 + n^2 = 2n^2 + 2n + 1. The whole figure has F(n) = 1 + 5 + ... + L(n-1) + L(n) + L(n-1) + ... + 5 + 1 = sigma_{i=1}^n L(n) + sigma_{i=1}^{n-1} L(n) cubes in it. To evaluate this you need to know that sigma_{i=1}^n n^2 = n(n+1)(2n+1)/6.

3. how can i get the nth term of the sequence : 7,25,63,129,231,377,575

i known the answer is 4n3+6n3+8n+3/3

how can i get the answer exactly like this?

4. Originally Posted by rpatel
how can i get the nth term of the sequence : 7,25,63,129,231,377,575

i known the answer is 4n3+6n3+8n+3/3

how can i get the answer exactly like this?
f(n)=an^3+bn^2+cn+d
You need to find the differences until you get the common difference.

1 7, 25, 63, 129, 231, 377, 575
6 18 38 66 102 146 198 1d
12 20 28 36 44 52 2d
8 8 8 8 8 3d
NB the underlined numbers are for reversing.
therefore to find a: 2a=3diff=8
therefore a = 4
To find d you need to find f(0) by revesing. therefore d= 1
to find b and c you need two equations, therefore the easiest is f(1) and f(2) by solving these equations silmultaneously you will find the desired values i.e b=-6 and c=8.