the series seems to have the nth term as n(n-1)/2 ; n>=3. So the 12th term is (12 x 11)/2 = 66.

That is false. The series starts $1,2,3,6,12,\ldots$, so already your formula fails at $n=5$. The general term is rather $a_n = 3\cdot 2^{n-3}$ for $n\geq 3$, easily proven by induction since $a_{n+1} = (1+2) + 3\sum_{k=0}^{n-3} 2^k = 3\cdot 2^{(n+1)-3}$. Therefore $a_{12} = 3\cdot 2^{9} = 1536$.

$a_{n+1} = (1+2) + 3\sum_{k=0}^{n-3} 2^k = 3\cdot 2^{(n+1)-3}$ can anyone explain this thing

$\sum_{k=0}^{n-3} 2^k = 2^{n-2}-1$ $3\sum_{k=0}^{n-3} 2^k = 3 \cdot 2^{n-2}-1$ $(1+2) + 3\sum_{k=0}^{n-3} 2^k = 3 \cdot 2^{n-2}$

If you named the first one $x$ it would be easy. Click to reveal hidden text

jedbear wrote: x,2x,4x,8x,16x,32x, 64x, 128x, 256x, 512x, 1024x, 2048x. This means the 12th is 2048. $4x\ne 2x+x$.......

Let the numbers be $a_1,a_2,a_3....$. We have $a_3=a_1+a_2, a_4=a_3+a_2+a_1=2(a_1+a_2), a_5= a_4+..+a_1=2(a_1+a_2)+2(a_1+a_2)=2^2(a_1+a_2)$ and so on. We can get from here the general term $a_n\not=a_1,a_2$ as $a_n=2^{n-3}(a_1+a_2)$ So the twelfth term in this series is $a_{12}=2^{12-3}(a_1+a_2)=2^9.3=1536.$

We have T1=1, T2=2 and T3 =3 . For n>3, We see , Tn = 2^(n-3)×T3 => T12 = 2^9×3 = 1536

Just use recurrence. $a_n = a_{n-1} + . . . + a_1$, with $a_1 = 1, a_2=2$. Then badda bing badda boom. You get $a_n = 3(2^{n-3})$. @mavropnevma's solution is nice. (I already knew the general formula for this recurrence but you could do it through case work, or maybe generating functions).