Can you unlock this pattern? n°3.

The terms are in the form $a_{n}=1^2+2^2+3^2.....+n^2$ , or in other words the sum of the first $n$ squares. So the $6^{th}$ term is the sum of the first $6$ squares, which is 204 .

5-1= 4 = (2*2)

14-5 = 9 = (3*3)

30-14 = 16 = (4*4)

55-30 = 25 = (5*5)

91-55=36 =(6*6)

140-91=49=(7*7)

so 140 + (8*8) = 204

The terms of this sequence are following this pattern:

$1=1^{2}$

$5=1^{2}+2^{2}$

$14=1^{2}+2^{2}+3^{2}$

$30=1^{2}+2^{2}+3^{2}+4^{2}$

$55=1^{2}+2^{2}+3^{2}+4^{2}+5^{2}$

$91=1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}$

$140=1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}+7^{2}$

So, the next number in the sequence is: $1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}+7^{2}+8^{2}=204$

Thus, the answer is $\boxed{204}$

nth term is Sqr of n + (n-1)th term

for the $n^{th}$ number in the sequence >> its value equal to : $\sum_{i=1}^n$ $n^2$

the difference is presented in the form of.......... 1^2+2^2+3^2..........and so on

sum of first n natural numbers........ so the number is 204

Use the form n^2, when n = 7, 140 + 8^2 = 204

1,(+4)5,(+9)14,.........,(+64) 201

The terms are in the form 1 1, 2 2-------

so 64 will be added to 140

The difference between the adjacent terms (n and n+1) is (n+1)^2 . Thus, the 8th term is 140 + (8)^2 = 204