1. Sequence

How do I find the general term a(n)?

2, 8, 18, 32, 50

I know a(n)= $2n^2$

But how do I get that?

I know you find the first order difference, b(n)

b(n)= 6, 10, 14, 18 = $4n-2$

$\sum_{k=1}^{n-1} 4k-2$

$=2+2n(n-1)-2(n-1)$

Obviously I messed up on that last step. How do I correctly do it? Thanks.

2. I don't really understood how you are doing but this is the way I do it

$S= 2+8+18+32+.....T_n$

$S= 0+2+8+18+....T_{n-1} + T_n$

Subtract the above two equations

$S-S = 2 +6+10+14+18+...+(T{n}-T_{n-1})~ - T_n$

$

T_n = 2 +6 +10+ 14+18+...(T{n}-T_{n-1})$

$

T_n= \text{Sum of the n terms of AP with first term 2 and common difference 4}$

$T_n = \frac{n(2\times 2+(n-1)4)}{2}$

$T_n=\frac{n(4n)}{2}$

$T_n= 2n^2$

-------------------------------------------------------------
we took first order difference of the complete sequence
and then the last term could be easily found by using sum of n terms of Arithmetic Progression

3. Oh, that's very clever!

chengbin, there is no "formula" for this- you just have try whatever you can think of and look for patterns. Since this sequence was neither an arithmetic sequence nor a geometric sequence, (the "easy" sequences), Adarsh tried looking at the "first difference" (the difference between consecutive terms) and saw that it was an arithmetic sequence (the "second difference" was constant).

I, seeing that all the numbers in the sequence are even, would have been inclined to divide each term, giving 1, 4, 9, 25, ... and recognized those as squares. Since $a_n/2= n^2$, $a_n= 2n^2$.