Question

A statement Sn about the positive integers is given. Write statements S1, S2, and S3, and show that each of these statements is true. Show your work.Sn:   1^2 + 4^2 + 7^2 + . . . + (3n - 2)^2 = 


Not exactly sure how to do this! Could someone help?

Answer 1

If T is each term in the sequence

T1 = (3*1-2)^2 = 1^2 = 1
T2 = (3*2-2)^2 = 4^2 = 16
T3 = (3*3-2)^2 = 7^2 = 49
T4 = (3*4-2)^2 = 10^2 = 100

then the sums are
S1 = 1
S2 = 1+16 = 17
S3 = 17 + 49 = 66
S4 = 66 + 100 = 166

Answer 2

S1: 1^2 = 1
S2: 1^2 + 4^2 = 17
S3: 1^2 + 4^2 + 7^2 = 66

You probably know that

1+2+3+... = n(n+1)/2
1^2 + 2^2 + 3^2 = n(n+1)(2n+1)/6
and so on

So, you'd expect your sum to be some kind of cubic polynomial

an^3 + bn^2 + cn + d

So, plug in your first four values:

a + b + c + d = 1
8a + 4b + 2c + d = 17
27a + 9b + 3c + d = 66
64a + 16b + 4c + d = 166

Solve to get

Sn = n/2 (6n^2-3n-1)

To prove it true for all k (not just k=1..4) use induction. Assume true for k:

S_{_k}1^2 + ... + (3k-2)^2 = k/2 (6k^2-3k-1)

Then, calculate for k+1:

S_{_k+1} = 1^2 + ... + (3k-2)^2 + (3(k+1)-2)^2
= 1^2 + ... + (3k-2)^2 + (9k^2+6k+1)
= k/2 (6k^2-3k-1) + 9k^2+6k+1
= 1/2 (6k^3+15k^2+11k+2)
= (k+1)/2 (6k^2+9k+2)
= (k+1)/2 (6(k+1)^2 - 3(k+1) - 1)

So, if true for n=k, it is true for n=k+1

Since it is true for n=1, it is true for n=2,3,4,... all values