In the sequence 1,8,15,22,29,36,43,50,57,64,...there occur a number of perfect squares (1,36,64...).?

Question

What would be the 100th Perfect Square in this such sequence.

I already know the answer but can YOU work it out!!

Answer 1

The series is 7n +1 and we wish to find the terms which are perfect squares, k^2

7n + 1 = k^2
7n = k^2 - 1 = (k+1) * (k-1)

Since 7 is a prime number, either k+1 or k-1 must be a multiple of 7 for this to work.

So k = 1 works
k = 6 and 8 work
k = 13 and 15 work

The (2i)th term that works is 7*i - 1
So for i = 50, k = 349
The 100th term is 349^2 = 121801

Answer 2

The first five are: 1: 1 2: 4 3: 9 4: 16 5: 25 As this passes 20 and 20 is not one of these, 20 cannot be a perfect square number

Answer 3

wait im mixed up
so add 7s
1+7=8+7=15+7=22+7=29+7=36+7=43+7=50+7=57+7=64...
then there is an add 5 add 2 sequence inside that sequence to get:
1+5=6+2=8+5=13+2=15+5=20+2=22+5=27+2=29...
then the 100th number is 349
oh okay i get it now so then you do 349*349=answer
i generated thousands of numbers and i got mixed up.
=(
whatever

Answer 4

It looks as if it is a sequence

1² , 6² , 8² , 13² , 15², 20² ...

which are two alternating APs

the 100th term will be the 50th term of the second AP

Un = (7n - 1)²

= 349²

121801

Answer 5

121801 = 349^2

As long as n^2 ≡ 1 (mod 7) it will be in the list.
This means n ≡ ±1 (mod 7)

1
6 and 8 squared (7±1)
13 and 15 squared (2*7±1)
20 and 22 squared (3*7±1)
etc...
349 and 351 squared (50*7±1)

349^2 is the 100th one.

Answer 6

the next squrare after 64 is (13x13) 169, the sequence is the original number + 7, Im not going all the way to the 100th though, i got to go to work!

Answer 7

121801

I just used about 6 lines of code. But I kept forgetting to include 1.

Answer 8

yup. Got it too.

Had to do it 'manually'. Curious how you would go about actually calculating that number.