thus
3^0 – 2^0 = 1 – 1 = 0
3^1 – 2^1 = 3 – 2 = 1
3^2 – 2^2 = 9 – 4 = 5
3^3 – 2^3 = 27 – 8 = 19
3^4 – 2^4 = 65
3^5 – 2^5 = 211
3^6 – 2^6 = 729 – 64 = 665

There is one more way to solve this

3*previous number + 2^(n-1)

0x3+1=1;
1×3+2=5;
5×3+4=19;
19×3+8=65;
65×3+16=211;
211×3+32=665;

wondering how it is possible?

lets solve 3*previous number + 2^(n-1)

=> 3*(3^(n-1) – 2^(n-1)) + 2^(n-1)
=> 3^n – (2+1)*2^(n-1) + 2^(n-1)
=> 3^n – 2^n – 2^(n-1) + 2^(n-1)
=> 3^n – 2^n i.e. our first equation

Rs 6 , Rs5 , Rs4, Rs3, Rs 2, and one item should be free Rs0 (free with other 4 items)

0+2+3+4+5+6=20

For each column

-2*2+3*3+5*5=30,
-3*3+4*4+6*6=43,
-1*1+4*4+6*6=51,
-2*2+5*5+7*7=70